Page 1 :
9. Arithmetic Progressions, Exercise 9.1, 1. Question, Write the first five terms of each of the following sequences whose nth terms are :, (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), Answer, (i), Put n = 1, A1 = 3(1) + 2 = 3+2=5, Put n = 2, A2 = 3(2) + 2=8, Put n = 3, A3 = 3(3) + 2 = 9+2=11, Put n = 4, A4 = 3(4) +2 = 12+2=14, Put n =5
Page 2 :
A5 = 3(5) + 2 = 15+2=17, (ii), Put n = 1, A1=, , =, , Put n = 2, A2=, , =0, , Put n = 3, A3=, , =, , Put n = 4, A4=, , =, , Put n = 5, A5=, , =, , =1, , (iii), Put n = 1, A1= 31= 3, Put n = 2, A2= 32= 9, Put n = 3, A3= 33= 27, Put n = 4, A4= 34= 81, Put n = 5, A5= 35= 243, (iv), Put n = 1, A1=, Put n = 2, , =, , =
Page 3 :
A2=, , =, , =, , =, , =, , Put n = 3, A3=, Put n = 4, A4=, , =, , =, , =2, , Put n = 5, A5 =, , =, , =, , (v), Put n = 1, A1= (-1)1.21 = (-1).2 = -2, Put n = 2, A2= (-1)2.22 = (1).4 = 4, Put n = 3, A3= (-1)3.23 = (-1).8 = -8, Put n = 4, A4= (-1)4.24 = (1).16 = 16, Put n = 5, A5= (-1)5.25 = (-1).32 = -32, (vi), Put n = 1, A1=, , =, , Put n = 2, A2=, , =0, , Put n = 3, A3=, Put n = 4, , =
Page 4 :
A4=, , =, , =4, , Put n = 5, A5=, , =, , (vii), Put n = 1, A1= (1)2 – 1+1 = 1, Put n = 2, A2= (2)2 -2+1 = 3, Put n = 3, A3= (3)2 -3+1 = 9-2 = 7, Put n = 4, A4 = (4)2 -4+1 = 16-3 = 13, Put n = 5, A5= (5)2 – 5+1 = 25-4 = 21, (viii), Put n = 1, A1= 2(1)2– 3(1) + 1 = 2-3+1 = 0, Put n = 2, A2= 2(2)2 – 3(2) + 1 = 8-6+1 = 3, Put n = 3, A3= 2(3)2 – 3(3) + 1 = 18-9+1 = 10, Put n = 4, A4= 2(4)2 – 3(4) + 1 = 32-12+1 = 21, Put n = 5, A5= 2(5)2 – 3(5) + 1 = 50-15+1 = 36, (ix), Put n = 1
Page 5 :
A1=, , =, , Put n = 2, A2=, , =, , Put n = 3, A3=, , =, , =, , Put n = 4, A4=, , =, , Put n = 5, A5=, , =, , 2. Question, Find the indicated terms in each of the following sequences whose nth terms are:, (i), (ii), (iii), (iv), (v), Answer, (i), Put n = 12, A12= 5(12) – 4, = 60-4 = 56, Put n = 15, A15= 5(15) – 4, = 75-4 = 71, (ii), Put n = 7
Page 6 :
A7=, , =, , =, , =, , =, , Put n = 8, A8=, (iii), Put n = 5, A5= 5(5-1) (5-2), = 5(4)(3) = 60, Put n = 8, A8= 8(8-1) (8-2), = 8(7)(6) = 336, (iv), Put n = 1, A1= 1(1-1) (2-1) (2-2) (3+2), =0, Put n = 2, A2= 2(2-1) (2-2) (3+2), =0, Put n = 3, A3= 3(3-1) (2-3) (3+3), = 3 (2) (-1) (6), = 36, (v), Put n = 3, A3= (-1)3 (3) = -3, Put n = 5, A5= (-1)5 (5) = -5, Put n = 8, A8= (-1)8 (8) = 8, 3. Question
Page 7 :
Find the next five terms of each of the following sequences given by:, (i), (ii), (iii), (iv), Answer, (i), Put n = 2, A2= a2-1 + 2, A2= a1 +2, = 1+2 = 3, Put n = 3, A3= a3-1 +2, = a2 + 2, = 3+2 = 5, Put n = 4, A4= a4-1 + 2, = a3 + 2, = 5+2 = 7, Put n = 5, A5= a5-1 + 2, = a4 + 2, = 7+2 = 9, Put n = 6, A6 = a6-1 +2, = a5 +2, = 9+2 = 11, (ii), Put n = 3
Page 10 :
For the following arithmetic progressions write the first term a and the common difference d :, (i), (ii), (iii), (iv), Answer, (i), First term, a = -5, Common difference, d = -1 – (-5), = -1 + 5 = 4, (ii), First term, a =, Common difference, d =, , -, , =, , (iii), First term, a = 0.3, Common difference, d = 0.55 – 0.30 = 0.25, (iv), First term, a = -1.1, Common difference, d = -3.1 – (-1.1), = -2, 2. Question, Write the arithmetic progression when first term a and common difference of d are as follows:, (i), (ii), (iii), Answer, (i), a1= 4, d= -3
Page 11 :
A2= a1+ d = 4-3 =1, A3= a1+ 2d = 4+ 2(-3) = 4-6 = -2, A4= a1 =3d = 4+ 3(-3) = 4-9 = -5, Series = 4, 1, -2, -5,…, (ii), a1= -1, d =, A2= a1 + d = -1 +, , =, , A3= a1 + 2d = -1 + 1 = 0, A4= a1 + 3d = -1 +, Series = -1,, , =, , , 0, ,…, , (iii), a1= -1.5, d= -0.5, A2= a1+ d = -1.5 – 0.5 = -2, A3= a1+ 2d = -1.5 + 2(-0.5) = -1.5 – 1 = -2.5, A4= a1+ 3d = -1.5 + 3(-0.5) = -1.5 – 1.5 = -3, Series =, 3. Question, In which of the following situation, the sequence of numbers formed will form an A.P.?, (i) The cost of digging a well for the first metre is Rs. 150 and rises by Rs.20 for each succeeding, metre., (ii)The mount of air present in the cylinder when a vacuum pump removes each time, remaining in the cylinder., Answer, (i) The cost of digging a well for the first metre = Rs. 150, The cost of digging a well for the second metre = 150 + 20 = Rs. 170, The cost of digging a well for the third metre = 170 + 20 = Rs. 190, The cost of digging a well for the fourth metre = 190 + 20 = Rs. 210, Difference between first and second metre = 170 – 150 = 20, , of their
Page 12 :
Difference between second and third metre = 190 – 170 = 20, Difference between third and fourth metre = 210 – 190 = 20, Since, all the differences are equal. Therefore, it’s an A.P., (ii) Let the amount of air present in the cylinder first time = x, Amount of air present in the cylinder second time = x -, , =, , Amount of air present in the cylinder third time =, , =, , -, , Difference in the amount of air present in the cylinder between first and the second time =, , Difference in the amount of air present in the cylinder between second and the third time =, =, Since, the differences are unequal. Therefore, it’s not an A.P., 4. Question, Show that the sequence defined by, , is an A.P., find its common difference., , Answer, Put n = 1, A1= 5(1) – 7, = -2, Put n = 2, A2= 5(2) – 7, = 10 – 7 = 3, Put n = 3, A3= 5(3) – 7, = 15 – 7 = 8, Common difference, d1= a2 – a1= 3 – (-2) = 5, Common difference, d2= a3 – a2= 8 – 3 = 5, Since, d1 = d2 and Common difference = 5, Therefore, it’s an A.P., 5. Question, , –x=, , -
Page 13 :
Show that the sequence defined by, , is not A.P., , Answer, Given: The sequence, , ., , To prove: the sequence defined by, Proof:Consider the sequence, , is not A.P., ,Put n = 1, , A1= 3(1)2 – 5 = 3 – 5 = -2, Put n = 2, A2= 3(2)2 – 5 = 12- 5 = 7, Put n = 3, A3= 3(3)2 – 5 = 27 – 5 = 22In an A.P the difference of consecutive terms should be same., So,, Common difference, d1= a2 – a1 =7 – (-2) = 9, Common difference, d2 = a3 – a2 = 22 – 9 = 13, Since, d1 ≠ d2, Therefore, it’s not an A.P., 6. Question, The general term of a sequence is given by, term and the common difference., Answer, Put n = 1, A1= -4(1)2 + 15 = 11, Put n = 2, A2= -4(2)2 + 15 = -1, Put n = 3, A3= -4(3)2 + 15 = -21, Common difference, d1= a2 – a1 = -1 – 11 = -12, Common difference, d2= a3 – a2 = -21 + 1 = 20, Since, d1 ≠ d2, Therefore, it’s not an A.P., , .Is the sequence an A.P..? If so, find its 15th
Page 14 :
7. Question, Find the common difference and write the next four terms of each of the following arithmetic, progressions:, (i), (ii), (iii), (iv), Answer, (i), Common difference, d = -2 – 1 = -3, A5= -8 – 3 = -11, A6= -11 – 3 = -14, A7= -14 – 3 = -17, A8= -17 – 3 = -20, (ii), Common difference, d = -3 – 0 = -3, A5= -9 – 3 = -12, A6= -12 – 3 = -15, A7= -15 – 3 = -18, A8= -18 – 3 = -21, (iii), Common difference, d =, A4=, A5=, , +, +, , A6= 4 +, A7=, , +, , =, =, , =4, , =, =, , =, , +1=
Page 16 :
(iv), Show that all of the above sequences form A.P., Answer, (i) Put n = 1, A1= 3 + 4(1) = 7, Put n = 2, A2= 3 + 4(2) = 11, Put n = 3, A3= 3 + 4(3) = 15, Common difference, d1= a2 – a1 = 11 – 7 = 4, Common difference, d2= a3 – a2 = 15 – 11 = 4, Since, d1 = d2, Therefore, it’s an A.P. with sequence 7, 11, 15,…, (ii) Put n = 1, A1= 5 + 2(1) = 7, Put n = 2, A2= 5 + 2(2) = 9, Put n = 3, A3= 5 + 2(3) = 11, Common difference, d1= a2 – a1= 9 – 7 = 2, Common difference, d2= a3 – a2= 11 – 9 = 2, Since, d1 = d2, Therefore, it’s an A.P. with sequence 7, 9, 11,…, (iii) Put n = 1, A1= 6 – 1 = 5, Put n = 2, A2= 6 – 2 = 4, Put n = 3, A3= 6 – 3 = 3
Page 17 :
Common difference, d1= a2 – a1= 4 – 5 = -1, Common difference, d2 = a3- a2= 3 – 4 = -1, Since, d1=d2, Therefore, it’s an A.P. with sequence 5, 4 , 3,…, (iv) Put n = 1, A1= 9 – 5(1) = 4, Put n = 2, A2= 9 – 5(2) = -1, Put n = 3, A3= 9 – 5(3) = -6, Common difference, d1= a2 – a1 = -1 – 4 = -5, Common difference,d2= a3 – a2 = -6 – (-1) = -5, Since, d1=d2, Therefore, it’s an A.P. with sequence 4, -1, -6,…, 10. Question, Find out which of the following sequences are arithmetic progressions. For those which are arithmetic, progressions, find out the common difference., (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), (x), (xi), (xii)
Page 18 :
Answer, (i), Common difference, d1 = 6 – 3 = 3, Common difference, d2= 12 – 6 = 6, Since, d1 ≠ d2, Therefore, it’s not an A.P., (ii), Common difference, d1 = -4 – 0 = -4, Common difference, d2= -8 – (-4) = - 4, Since, d1 = d2, Therefore, it’s an A.P. with common difference, d = -4, (iii), Common difference, d1=, Common difference, d2 =, , -, , =, =, , Since, d1 ≠ d2, Therefore, it’s not an A.P., (iv), Common difference, d1= 2 – 12 = -10, Common difference, d2= -8 -2 = -10, Since,d1 = d2, Therefore, it’s an A.P. with common difference, d = -10, (v), Common difference, d1= 3 – 3 = 0, Common difference, d2= 3 – 3 = 0, Since, d1=d2, Therefore, it’s an A.P. with common difference, d = 0, (vi), Common difference, d1= p + 90 – p = 90
Page 19 :
Common difference, d2= p + 180 – p – 90 = 90, Since, d1=d2, Therefore, it’s an A.P. with common difference, d = 90, (vii), Common difference, d1= 1.7 – 1.0 = 0.7, Common difference, d2= 2.4 – 1.7 = 0.7, Since, d1=d2, Therefore, it’s an A.P. with common difference, d = 0.7, (viii), Common difference, d1= -425 + 225 = -200, Common difference, d2= -625 + 425 = -200, Since, d1=d2, Therefore, it’s an A.P. with common difference, d = -200, (ix), Common difference, d1= 10 + 26 – 10 = 26 = 64, Common difference, d2 = 10 + 27 – 10 – 26 = 26 (2 – 1) = 64, Since, d1=d2, Therefore, it’s an A.P. with common difference, d = 64, (x), Common difference, d1 = (a + 1) + b – a – b = 1, Common difference, d2 = (a + 1) + (b + 1) – (a + 1) – b = 1, Since, d1 = d2, Therefore, it’s an A.P. with common difference, d = 1, (xi), Common difference, d1= 32 – 12 = 8, Common difference, d2 = 52 – 32 = 25 – 9 = 16, Since, d1≠d2, Therefore, it’s not an A.P., (xii)
Page 20 :
Common difference, d1 = 52 – 12 = 24, Common difference, d2 = 72 – 52 = 24, Since, d1 = d2, Therefore, it’s an A.P. with common difference, d = 24, 11. Question, Find the common difference of the A.P. and write the next two terms:, (i), (ii), (iii), (iv), (v), Answer, (i), Common difference, d = 59 – 51 = 8, A5 = 75 + 8 = 83, A6 = 83 + 8 = 91, (ii), Common difference, d = 67 – 75 = -8, A5= 51 - 8 = 43, A6= 43 – 8 = 35, (iii), Common difference, d = 2.0 – 1.8 = 0.2, A5= 2.4 + 0.2 = 2.6, A6= 2.6 + 0.2 = 2.8, (iv), Common difference, d =, A5=, , +, , =4, , –0=
Page 21 :
A6= 4 +, , =, , (v), Common difference, d = 136 – 119 = 17, A5= 170 + 17 = 187, A6= 187 + 17 = 204, 12. Question, The nth term of an A.P. is 6n+2. Find the common difference., Answer, an= 6n + 2, Put n = 1, A1= 6(1) + 2 = 8, Put n = 2, A2= 6(2) + 2 = 14, Common difference, d = a2 – a1 = 14 – 8 = 6, , Exercise 9.3, 1. Question, Find:, (i) 10th term of the A.P., (ii) 18th term of the A.P., (iii) nth term of the A.P., (iv) 10th term of the A.P., (v) 8th term of the A.P., (vi) 11th term of the A.P., (vii) 9th term of the A.P., Answer, (i) 10th term of the A.P., a = 1, d = 4 – 1 = 3, A10= a + (10 – 1) d
Page 22 :
= 1 + (9)3 = 28, (ii) 18th term of the A.P., a = √2, d = 3√2 - √2 = 2√2, A18= a + (18 – 1) d, = √2 + (17) 2√2 = 35√2, (iii) nth term of the A.P., a = 13, d = -5, An= a + (n-1)d = 13 + (n-1) (-5), = 13 -5n + 5 = 18 – 5n, (iv) 10th term of the A.P., a = -40, d = 25, A10= a + (10-1) d = -40 + 9 (25), = -40 + 225 = 185, (v) 8th term of the A.P., a = 117, d = -13, A8= a + (8 – 1) d = 117 + 7 (-13), = 117 – 91 = 26, (vi) 11th term of the A.P., a = 10.0, d = 10.5 – 10.0 = 0.5, A11= a + (11 – 1) d = 10.0 + 10 (0.5), = 10.0 + 5 = 15.0, (vii) 9th term of the A.P., a= ,d=, , -, , =, , A9= a + (9 – 1) d =, , +8( )=, , 2 A. Question, (i) Which term of the A.P., Answer, a = 3, d = 5, , is 248?
Page 23 :
An= a + (n – 1) d, 248 = 3 + (n – 1) 5 = 3 + 5n – 5, 248 = -2 + 5n, 250 = 5n, n = 50, 2 B. Question, Which term of the A.P., , is 0?, , Answer, a = 84, d = -4,An = 0, We know,An= a + (n – 1) d, 0 = 84 + (n – 1) -4, 0 = 84 – 4n + 4, 0 = 88 – 4n, n = 22, 2 C. Question, Which term of the A.P., , is 254?, , Answer, a = 4, d = 5, An= a + (n – 1) d, 254 = 4 + (n – 1) 5, 254 = 4 + 5n – 5, 254 = -1 + 5n, 255 = 5n, n = 51, 2 D. Question, Which term of the A.P., Answer, a = 21, d = 21, An= a + (n – 1) d, 420 = 21 + (n – 1) 21, , is 420?
Page 24 :
420 = 21 + 21n – 21, 420 = 21n, n = 20, 2 E. Question, Which term of the A.P., , is its first negative term?, , Answer, a = 121, d = -4, An = -ve, An= a + (n – 1) d, = 121 + (n – 1) -4, = 125 – 4n, Since, we want –ve term, Therefore, 4n should be a number greater than 125, Hence, 128 is the number greater than 125 and divisible by 4, 4n = 128, n = 32, Therefore, 32nd term of the given A.P. is –ve, 3 A. Question, Is 68 a term of the A.P., , ?, , Answer, a = 7, d = 3, An= a + (n – 1) d, 68= 7 + (n – 1) 3, 68= 4 + 3n, 3n = 64, Since, 64 is not divisible by 3, Therefore, 68 isn’t the term of the given A.P., 3 B. Question, Is 302 a term of the A.P., Answer, , ?
Page 25 :
a = 3, d = 5, An= a + (n – 1) d, 302 = 3 + (n – 1) 5, 302 = -2 + 5n, 304 = 5n, Since, 304 is not divisible by 5, Therefore, 302 isn’t the term of the given A.P., 3. Question, Is -150 a term of the A.P., , ?, , Answer, a = 11, d = -3, An= a + (n – 1) d, -150 = 11 + (n – 1) (-3), -150 = 11 – 3n + 3, -150 = 14 – 3n, -164 = 3n, Since, -164 is not divisible by 3, Therefore, -150 isn’t the term of the given A.P., 4. Question, How many terms are there in the A.P.?, (i), (ii), (iii), (iv), Answer, (i), a = 7, d = 3, An = 43, a + (n – 1) d = 43
Page 26 :
7 + (n – 1) 3 = 43, 3n = 39, n = 13, Therefore, there are total 13 terms in the A.P., (ii), a = -1, d =, An =, a + (n – 1) d =, -1 + (n -1), , =, , n = 26, (iii), a = 7, d = 6, An = 205, a + (n – 1) d = 205, 7 + (n – 1) 6 = 205, 6n = 204, n = 34, (iv), a = 18, d =, An = -47, a + (n – 1) d = -47, 18 + (n – 1), 18 -, , +, , = -47, , = -47, , n = 27, 5. Question, The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of, terms.
Page 27 :
Answer, Given: The first term of an A.P. is 5, the common difference is 3 and the last term is 80To find: the, number of terms.Solution :We have a = 5, d = 3, an = 80From the formula an = a + (n-1) d Number of terms is :, , =, =, , +1, +1, , = 25 + 1, = 26Hence the number of terms in a given A.P. is 26., 6. Question, The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term., Answer, a6= 19 = a + (n – 1) d, =19 = a + 5d …(i), A17 = 41 = a + (n – 1) d, = 41 = a + 16d …(ii), Subtracting (i) from (ii), we get, 22 = 11d, d=2, Now substituting the value of d in (i): 19 = a + 10, =a=9, A40= a + (40 – 1) 2, = 9 + 78, = 87, 7. Question, If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term., Answer, a9 = 0, a + (9 – 1)d =0, a + 8d = 0
Page 30 :
⇒ a = md- 2nd, , (i), , LHS: a3m+1= a + (3m +1 -1)d = md – 2nd + 3md = 2d(2m - n), RHS: 2a(m + n + 1) = 2[a +(m +n +1 -1)d]= 2[md -2nd +md +nd] = 2d(2m - n), LHS = RHS, Hence, proved, 12. Question, If the nth term of the A.P., , is same as the nth term of the A.P., , Answer, A=9, D=7-9=-2, a=15, d=-3, An=an, A+(n -1)D=a +(n -1)d, 9 – (n -1)2=15 – (n -1)3, (n -1)(3 -2)=6, n -1=6, n=7, 13. Question, Find the 12th term from the end of the following arithmetic progressions:, (i), (ii), (iii), Answer, (i), a=3, d=5 -3=2, an =201, an= a +(n -1)d, 201 =3 +2n – 2, N =100, Now, we have to find 12th term from the last that means,, 100th – 11=89th term, Then,, , find n.
Page 31 :
a89 =a + (89 – 1)d, =3 +88, =179, Hence, the 12th term from the end of the A.P. is 179., (ii), a=3, d= 8 -3 =5, an= 253, a + (n -1)d =253, 3 + (n -1)5 =253, n=51, Now, we have to find 12th term from the last that means,, 51th -11 = 40th term, Then,, a40 = a + (n -1)d, = 3 + 39(5), =198, Hence, the 12th term from the end of the A.P. is 198, (iii), a=1, d= 4 -1 =3, an= 88, a + (n -1)d =88, 1 + (n -1)3 =88, n =30, Now, we have to find 12th term from the last that means,, 30th -11 = 19th term, a19 = a + 18d, = 1 + 18(3), =55, Hence, the 12th term from the end of the A.P. is 198.
Page 33 :
Now, difference, d=, , ⇒ d= 2, Hence, first term is 3 and the common difference is 2., 15. Question, Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22., Answer, a6= a + 5d, 12= a + 5d …(i), a8= a + 7d, 22= a + 7d …(ii), Subtracting (i) from (ii), we get,, 10= 2d, d= 5, from (i), 12 = a + 5(5), 12= a + 25, a= -13, a2= a +d, =-13 +5, =-8, an= a + (n-1)d, = -13 + (n-1)5, =-18 +5n, 16. Question, How many numbers of two digit are divisible by 3?, Answer
Page 34 :
Two digit numbers divisible by 3 are 12, 15, 18,… ,99, Hence, a= 12, ,d=3, an= a + (n -1)d, 99= 12(n -1)3, 99= 9 = 3n, n=30, Hence, the total numbers of two digit divisible by 3 are 30., 17. Question, An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term., Answer, n =60, a=7, a60 =l =125, a60 = a + 59d, 125 = 7 + 59d, d=, d=2, a32= a + 31d, = 7 + 31(2), =69, Hence, 32th term is 69 in the given A.P., 18. Question, The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the, first term and the common difference of the A.P.?, Answer, Given: a4 + a8 = 24 …(i), a6 + a10 = 34 …(ii), a4 = a + 3d, a8 =a + 7d, a6 =a + 5d, a10=a + 9d, Put the value of a4 and a8 in (i)
Page 35 :
(a + 3d) + (a + 7d) = 24, a + 5d =12 …(iii), Put the value of a6 and a10 in (ii), (a +5d) + (a +9d) =34, a +7d =17 …(iv), Subtracting (iii) from (iv), we get, 2d = 5, d=, Putting the value of d in (iii), we get, a = 12 -, , =, , Hence, first term is, , and common difference is, , 19. Question, The first term of an A.P. is 5 and its 100th term is - 292. Find the 50th term of this A.P., Answer, Given, a = 5, a100= -292, 5 – 99d = -292, d=, , =3, , Now, 50th term, a50 = a + (50 – 1) d, = 5 + (49) 3 = 5 + 147 = 152, Hence, 50th term of given A.P. is 152, 20. Question, Find, , for the A.P., , (i), (ii), Answer, (i), a = -9, d = -14 + 9 = -5
Page 37 :
a13 = 79, a + 12d = 79 (ii), By subtracting (i) from (ii), we get, 2d = 74, d = 37, Hence, the common difference is 37, (ii)a10 = a + 9d (i), a5 = a + 4d (ii), a10 – a5 = a + 9d – a – 4d, 200 = 5d, d = 40, Hence, common difference is 40, (iii) a20 = a + 19d (i), a18 = a + 17d (ii), Given that a20 = a18 + 10, a + 19d = a + 17d + 10, 2d = 10, d=5, Hence, common difference is 5, 22. Question, Find n if the given value of x is the nth term of the given A.P., (i), (ii), (iii), (iv), Answer, (i), a = 25, d = 50 – 25 = 25, Last term, l = 1000
Page 38 :
Number of terms, n =, =, , +1, , + 1 = 40, , Hence, the value of n is 40, (ii), a = -1, d = -2, Last term, l = -151, Number of terms, n =, =, , +1, , + 1 = 76, , Hence, the value of n is 76, (iii), a=, , ,d=, , Last term, l = 550, l = a + (n – 1) d, 550 =, , + (n – 1), , 550 =, 1100 = 11n, n = 100, Hence, number of terms, n = 100, (iv), a = 1, d =, , –1=, , Last term, l =, a + (n – 1) d =, 1 + (n – 1), =, , =, , –1+
Page 39 :
=, 10n = 170, n = 17, Hence, value of n is 17, 23. Question, If an A.P. consists of n terms with first term a and nth term ls how that the sum of the mth term from, the beginning and the mth term from the end is (a+l)., Answer, Given, a = a, Last term, l = l, We have to prove that the sum of the mth term from the beginning and mth term from the end is (a +, 1), Now, mth term from the beginning,, am(b) = a + (m – 1) l, = a + ml – l (i), Again mth term from the end, am(e) = l – (m – 1)l, = l – ml + l = 2l – ml (ii), By adding (i) and (ii), we get, a + ml – l + 2l – ml = a + l, Hence, proved, 24. Question, Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12., Answer, Given, a3 = 16, a + 2d = 16 (i), a5 = a + (5 – 1) d, = a + 4d, a7 = a + (7 -1)d, = a + 6d
Page 40 :
Acc. To question,, a7 = 12 + a5 (ii), Put the value of a5 and a7 in (ii), we get, a + 6d = 12 + a + 4d, 2d = 12 + a – a, 2d = 12, d=6, From equation (i),, 16 = a + 2(6), 16 = a + 12, a=4, We have to find A.P., a1 = 4, a2 = a + d = 4 + 6 = 10, a3 = a + 2d = 4 + 2(6) = 16, a4 = a + 3d = 4 + 3(6) = 22, Hence, the A.P. is 4 , 10, 16 , 22,…, 25. Question, The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P., Answer, a7 = 32, a + 6d = 32 (i), a13 = a + 12d = 62 (ii), By subtracting (i) from (ii), we get, 6d = 30, d= 5, From (i), a = 32 – 6(5) = 2, We have to find A.P., a1 = 2, a2 = a + d = 2 + 5 = 7
Page 41 :
a3 = a + 2d = 2 + 2(5) = 12, a4 = a + 3d = 2 + 3(5) = 17, Hence, A.P. is 2, 7, 12, 17,…, 26. Question, Which term of the A.P., , will be 84 more than its 13th term?, , Answer, a = 3, d = 7, a13 = 3 + (13 – 1)7, = 3 + 84 = 87, Now, nth term is 84 more than the 13th term, an = 84 + a13, = 84 + 87, = 171, Now we have to find n,, an = a + (n – 1) d, 171 = 3 + (n – 1) 7, 7n = 175, n = 25, Hence, 25th term of the given A,.P. is 84 more than its 13th term, 27. Question, Two arithmetic progressions have the same common difference. The difference between their 100th, terms is 100, what is the difference between their 1000th terms?, Answer, Two arithmetic progressions have the same common difference., Let, common difference = d, First term of first A.P. = a, First term of second A.P. = a’, Given that difference between their 100th term is 100, 100th term of first A.P., a100 = a + 99d, 100th term of second A.P., a’100 = a’ + 99d
Page 42 :
Acc. To question,, a100 – a’100 = 100, a + 99d – a’ – 99d = 100, a – a’ = 100 (i), 1000th term of first A.P., a1000 = a + 999d, 1000th term of second A.P., a’1000 = a’ + 999d, Acc. To question,, a1000 – a’1000 = a + 999d – a’ – 999d, = a – a’, a1000 – a’1000 = 100 [from (i)], Hence, difference between 1000th term of two A.P. is 100, 28. Question, For what value of n, the nth terms of the arithmetic progressions, Answer, Given, A.P. = 63, 65, 67,…, A.P.’ = 3, 10, 17,…, Let for A.P.,, First term, a = 63, d = 65 – 63 = 2, And nth term = an, an = 63 + (n – 1) 2, = 63 + 2n – 2, = 61 + 2n, For A.P.’, First term, a = 3, d = 10 – 3 = 7, nth term = an, Now nth term of both A.P. are equal, 61 + 2n = -4 + 7n, 7n – 2n = 61 + 4, n = 13, , and, , are equal?
Page 43 :
Hence, 13th term of both the A.P. are equal, 29. Question, How many multiples of 4 lie between 10 and 250?, Answer, Let, multiple of 4 lies between 10 and 250, 12, 16, 20, 24,…248, We know, an = a + (n – 1)d, First term, a = 12, C0mmon difference, d = 16 – 12 = 4, Last term, an = 248, an = a+ (n – 1)d, 248 = 12 + (n – 1) 4, 248 = 12 + 4n – 4, 4n = 248 – 12 + 4, 4n = 240, n = 60, Hence, multiple of 4 lies between 10 and 250 is 60, 30. Question, How many three digit numbers are divisible by 7?, Answer, Let, three digit number divisible by 7 are, 105, 112, 119,….994, Here, First term, a = 105, Common difference, d = 112 – 105 = 7, And last term, an = 994, an = a+ (n – 1) d, 994 = 105 + (n – 1) 7, 994 = 105 + 7n – 7, 994 = 98 + 7n, 7n = 896
Page 44 :
n = 128, Hence, three digit number that are divisible by 7 are 128, 31. Question, Which term of the arithmetic progression, , will be72 more than its 41th term?, , Answer, a = 8, d = 6,, Last term = an, an= a + (n – 1) d, = 8 + (n – 1) 6, = 2 + 6n (i), Let, 41st term, a41 = 8 + (41 – 1) 6, = 8 + 40 * 6 = 248, Now the term is 72 more than its 41st term, an = 72 + a41, = 72 + 248 = 320, Putting this value in (i), we get, 320 = 2 + 6n, 6n = 318, n = 53, Hence, 53rd term of the given A.P. is 72 more than its 41st term, 32. Question, Find the term of the arithmetic progression, Answer, First term, a = 9, d = 12 – 9 = 3, Let, last term be an, an = a + (n – 1) d, = 9 + (n – 1)3, = 9 + 3n – 3 = 6 + 3n (i), Let, 36th term, a36 = a + 35d, = 9 + 35 (3) = 114, , which is 39 more than its 36th term.
Page 45 :
Now the term is 39 more than its 36th term, an = 39 + a36, = 39 + 114 = 153, Putting the value in (i), we get, 153 = 6 + 3N, 3n = 153 – 6, 3n = 147, n = 49, Hence, 49th term of the given A.P. is 39 more than its 36th term, 33. Question, Find the 8th term from the end of the A.P., Answer, a = 7, d = 3, Last term, an = 184, an = a + (n – 1) d, 184 = 7 + (n - 1) 3, 184 = 7 + 3n – 3, 180 = 3n, n = 60, Now we have to find last 8th term, it means = 60th – 7 = 53rd term, a53 = 7 + (53 – 1)3, = 7 + 52 * 3, = 7 + 156 = 163, Hence, 8th term from the end of given is 163, 34. Question, Find the 10th term from the end of the A.P. 8, 10, 12, ..., 126, Answer, Given,First term, a = 8Common difference, d = 2, Let the number of terms be 'n'We know, nth term of an AP isan = a + (n – 1)dwhere ‘a’ and ‘d’ are, first term and common difference of AP respectivelyLast term, an = 126
Page 46 :
⇒ an = a + (n – 1) d, ⇒ 126 = 8 + (n – 1) 2, ⇒ 120 = 2n, ⇒ n = 60, Now we have to find last 10th term, means = 60th term – 10 + 1 = 51th term, Now, a51 = 8 + (51 – 1)2, = 8 + 50(2) = 8 + 100= 108, Hence, 10th term from the last of given A.P. is 108, 35. Question, The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P., Answer, Given, a4 + a8 = 24 (i), a6 + a10 = 44 (ii), We know, an = a + (n – 1) d, 4th term, a4 = a + 3d, 6th term, a6 = a + 5d, 8th term, a8 = a + 7d, 10th term, a10 = a + 9d, Putting the value of a4 and a8 in (i), we get, a + 3d + a + 7d = 24, 2a + 10d = 24 (iii), Put the value of a6 and a10 in (ii), we get, a + 5d + a + 9d = 44, 2a + 14d = 44 (iv), By subtracting (iii) from (iv), we get, 4d = 20, d=5, Now putting value of d in (iii), we get, 2a = 24 – 10(5)
Page 47 :
a = -13, a1 = -13, a2 = a + d = -13 + 5 = - 8, a3 = a + 2d = -13 + 10 = -3, Hence, the A.P. is -13, -8, -3,…, 36. Question, Which term of the A.P., , will be 120 more than its 21st term?, , Answer, a = 3, d = 15 – 3 = 12, Let last term be an, an = a + (n – 1) d, = 3 + (n – 1) 12, = 12n – 9 (i), a21 = a + 20d, = 3 + 20(12) = 243, Now, the term is 120 more than the 21st term, an = 120 + a21, = 120 + 243, = 363, Putting this value in (i), we get, 363 = 12n – 9, 12n = 363 + 9, n = 31, Hence, 31st term of given A.P. is 120 more than its 21st term, 37. Question, The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the, nth term., Answer, Given: The 17th term of an A.P. is 5 more than twice its 8th term and the 11th term of the A.P. is 43., To find: the nth term.
Page 49 :
=, =, , +1, +1, , =100, Hence, the number of all three digit natural numbers which are divisible by 9 are 100, 39. Question, The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P., Answer, Given: a9 =19, a + 8d= 19 (i), Acc. to question,, a19= 3a6 (ii), a19=a + 18d, a6=a + 5d, Putting the value of a19 and a6 in (ii), a + 18d =3(a +5d), 3d= 2a, 3d=2(19 -8d) (from (i)), 19d=38, d=2, Now, putting the value of d in (i), a=19 -8(2) =3, a1 = a=3, a2 =a +d=3 +2=5, a3 =a +2d=3 + 2(2)=7, Hence, the A.P. is, 40. Question, The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P., Answer, Given: a5 = 22
Page 51 :
=4a +56d, =4(5d) +56d [from (i)], =20d +56d, =76d, Since, L.H.S=R.H.S., Hence proved, 42. Question, Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5., Answer, We have to find the numbers between 101 and 999 which are divisible by both 2 and 5, That means the numbers divisible by 10 between 101 and 999., Let, the required divisible numbers be,, 110, 120, 130, …, 990, a=110, d=10, last term, l=990, Number of terms, n =, =, =, , +1, , +1, +1, , =89, Hence, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 are, 89., 43. Question, If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)rd term., Answer, We know, nth term of an AP isan = a + (n – 1)dwhere ‘a’ and ‘d’ are first term and common difference, of AP respectively, Given,
Page 52 :
Subtracting equation [1] from [2], we get, , Putting this value of d in equation [1], we get, , Now, 63rd term = a + 62d, , Hence Proved!
Page 53 :
44. Question, The sum of 5th and 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP., Answer, a5 +a9=30, (a +4d) + (a +8d)=30, 2a +12d=30, a + 6d=15 (i), Acc. To question,, a25 =3a8, a+ 24d = 3(a +7d), 3d=2a, 3d=2(15 -6d) [from(i)], 15d=30, d=2, Putting the value of d in (i),, a= 15 – 6(2)=3, a1 = a=3, a2 =a +d=3 +2=5, a3 =a +2d=3 + 2(2)=7, Hence, the A.P. is, 45. Question, Find where 0 (zero) is a term of the A.P., Answer, a=40, d=37 -40= -3, Let, an =0, a + (n -1) d =0, 40 + (n -1) (-3) =0, -3n = -43, n=43/3, Since, ‘n’ can’t be a fraction
Page 54 :
Hence, the answer is no, 46. Question, Find the middle term of the A.P., , ., , Answer, a=213, l=37, Middle term of A.P. =, =, =, =125, 47. Question, If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P., Answer, a5=31, a +4d =31 (i), a25= 140 + a5, a + 24d= 140 + 31, 31 -4d +24d=171 [ from (i) ], 20d=140, d=7, Putting the value of d in (i), a= 31 – 4(7) = 3, a1 = a=3, a2 =a +d=3 +7=10, a3 =a +2d=3 + 2(7) =17, Hence, the A.P. is 3, 10, 17,…, , Exercise 9.4, 1. Question, The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds these, term by 6, find three terms.
Page 55 :
Answer, Let the three terms be a - d, a, a + d, a – d + a + a – d = 21 (given), 3a = 21, a=7, According to question,, (a – d) (a + d) = a + 6, a2 – d2 = a + 6, 72 – d2 = 7 + 6, 49 – d2 = 13, d2 = 36, d=6, Therefore, a – d = 7 – 6 = 1, a + d = 7 + 6 = 13, Thus, the three terms are 1, 7 and 13., 2. Question, Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers., Answer, Let the three numbers be a – d, a, a + d, a – d + a + a + d = 27, 3a = 27, a=9, According to question,, (a – d) (a) (a + d) = 648, (a2 – d2) (a) = 648, (92 – d2) 9 = 648, (81 – d2) = 72, d2 = 9, d=3, Therefore, a – d = 9 – 3 = 6
Page 56 :
A + d = 9 + 3 = 12, 3. Question, Find the four numbers in A.P. whose sum is 50 and in which the greatest number is 4 times the least., Answer, Let the four numbers be (a – 3d), (a – d), (a + d) and (a + 3d), a – 3d + a – d + a + d + a + 3d = 50, 4a = 50, a=, According to question,, (a + 3d) = 4(a – 3d), a + 3d = 4a – 12d, 3a = 15d, 5d =, d=, Therefore,, a – 3d =, , -, , a–d=, , -, , a+d=, , +, , a + 3d =, , =5, = 10, , +, , = 15, = 20, , 4. Question, The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles., Answer, Let the angles be (a – 3d), (a – d), (a + d) and (a + 3d), a – 3d + a – d + a + d + a + 3d = 360o (sum of angles of quadrilateral), 4a = 360o, a = 90o, According to question,
Page 59 :
d2 = 2ab, Since, d1 = d2 i.e. the common difference is same., Therefore, the given terms are in A.P., , Exercise 9.5, 1. Question, Find the sum of the following arithmetic progressions:, (i), , to 10 terms, , (ii), , to 12 terms, , (iii), , to 25 terms, , (iv), , to 12 terms, , (v), , to 22 terms, , (vi), , to n terms, , (vii), , to n terms, , (viii), , to 36 terms, , Answer, (i), , to 10 terms, , Sn =, , [2a + (n – 1) d], , S10 =, , [100 + 9(-4)], , = 5 [100 – 36], = 5(64) = 320, (ii), , to 12 terms, , Sn =, =, , [2a + (n – 1) d], [2 + (12 – 1) 2], , = 6 (2 + 22), = 144, (iii), Sn =, , to 25 terms, [2a + (n – 1) d]
Page 60 :
=, , [2 (3) + 24( )], , =, , [6 + 36], , = 525, (iv), , to 12 terms, , Sn =, =, , [2a + (n – 1) d], [82 - 55], , = 6 [27] = 162, (v), , to 22 terms, , Sn =, =, , [2a + (n – 1) d], [2(a + b) + (21) (-2b)], , = 11 [2a + 2b – 42b], = 11 [2a – 40b], = 22a – 440b, (vi), Sn =, , to n terms, [2a + (n – 1) d], , =, , [2(x – y)2 + (n – 1) (2xy)], , =, , (2) [(x – y)2 + (n – 1) xy], , = n [(x – y)2 + (n – 1) xy], (vii), Sn =, =, , [2(, , to n terms, [2a + (n – 1) d], ) + (n – 1) xy], , =, , {2(x – y) + (n – 1) (2x – y)}, , =, , {n (2x – y) – y}, , (viii), Sn =, , to 36 terms, [2a + (n – 1) d]
Page 61 :
=, , [-52 + (35) 2], , = 18 [-52 + 70], = 18 [18] = 324, 2. Question, Find the sum on n term of the A.P., Answer, Sn =, , [2a + (n – 1) d], , =, , [10 + (n – 1) -3], , =, , [13 – 3n], , 3. Question, Find the sum of n terms of an A.P. whose nth term is given by, Answer, Put n = 1, a1 = 5 – 6(1) = 5 – 6 = -1, Put n = 2, a2= 5 – 6(2) = 5 – 12 = -7, a = -1, d = -7 + 1 = -6, Sn =, , [2a + (n – 1) d], , =, , [2(-1) + (n -1) (-6)], , =, , [-1 + 3n + 3], , = n [2 -3n], 4. Question, If the n sum of a certain number of terms starting from first term of an A.P. is, the last term., Answer, a = 25, d = -3, Sn =, 116 =, , [2a + (n – 1) d], [50 – 3n +3], , is 116. Find
Page 62 :
116 =, , [53 – 3n], , 232 = 53n – 3n2, 3n2 – 53n + 232 = 0, n = 8 or n =, , , which isn’t possible as n must be a natural number., , Therefore, n = 8, [a + l] = 116, [25 + l] = 116, , l=4, Hence, the last term is 4, 5 A. Question, How many terms of the sequence, , should be taken so that their sum is zero?, , Answer, Sn =, 0=, , [2a + (n – 1) d], [2(18) + (n – 1) (-2)], , 38n – 2n2 = 0, n = 0 (which is not possible), n = 19, Therefore, n = 19 terms, 5 B. Question, How many terms are there in the A.P. whose first and fifth terms are 14 and 2 respectively and the, sum of the terms is 40?, Answer, a5= a + (n -1) d, 2 = -14 + 4d, 16 = 4d, d=4, Sn =, , [2a + (n – 1) d]
Page 63 :
40 =, , [2(-14) + (n -1) (4)], , 40 = 12n2 – 16n, 2n2 – 16n -40 = 0, n2 – 8n -20 = 0, n2 – 10n + 2n – 20 = 0, (n – 10) (n + 2) = 0, n = 10 and n = -2 (not possible), 5 C. Question, How many terms of the A.P., , must be taken so that their sum is 636?, , Answer, Sn =, , [2a + (n – 1) d], , 636 = [2(9) + (n – 1) 8], 636 = n [9 + 4n -4], 636 = 5n + 4n2, 4n2 + 5n – 636 = 0, n = 12 or n =, , (not possible), , 5 D. Question, How many terms of the A.P., , must be taken so that their sum is 693?, , Answer, Sn =, , [2a + (n – 1) d], , 693 =, , [2(63) + (n – 1) (-3)], , 693 =, , [126 -3n + 3], , 462 = n [43 – n], n2 – 43n + 462 = 0, n = 22 or n = 21, 6. Question, The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how, many terms are there and what is their sum?
Page 64 :
Answer, a = 17, l = 350, d = 9, Number of terms, n =, =, , +1, , =, , =, , Sn=, =, , +1, , = 38, , (a + l), (350 + 17), , = 19 (367) = 6973, 7. Question, The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the, first term, the common difference and the sum of first 20 terms., Answer, We know, nth term of an AP isan = a + (n – 1)dwhere ‘a’ and ‘d’ are first term and common difference, of AP respectivelyGiven, third term of an A.P. is 7a3 = 7⇒ a + 2d = 7, (i)the seventh term exceeds, three times the third term by 2, ⇒ a7 = 3a3 + 2⇒ a + 6d = 3(7) + 2⇒ 7 – 2d +6d = 21 + 2⇒ 4d = 16⇒ d = 4, From (i), a = 7 – 2(4), = -1, Also, we know sum of first 'n' terms of an AP is, , Sum of first 20 terms is, , = 10[-2 + 76]= 740, 8. Question, The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the, common difference., Answer, a = 2, l = 50, Sn=, , [a + l], , 442 =, , [50 +2], , n=, , = 17
Page 65 :
n=, , +1, , 17 =, , +1, , 16d = 48, d=3, 9. Question, If 12th term of an A.P. is 13 and the sum of the first four terms is 24, what is the sum of first 10, terms?, Answer, a12= -13, a + (12 – 1) d = -13, a + 11d = -13, S4 =, , [2a + (4 – 1) d], , 24 = 2 [2a + 3d], 12 = 2a + 3d, 12 = 2 (-11d – 13) + 3d, 12 = -22d – 26 + 3d, 38 = -19d, d = -2, a = -13 + 11(2), = -13 + 22, =9, S10=, , [2(9) + (10 – 1) (-2)], , = 5 [18 – 18], =0, 10. Question, Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149., Answer, a22= a + 21d, 149 = a + 21(22)
Page 66 :
a = 149 – 462, a = -313, S22=, , [2a + (22 – 1) d], , = 11 [-626 + 462], = 11 (-164), = -1804, 11. Question, Find the sum of all natural numbers between a and 100, which are divisible by 3., Answer, a = 3, l = 99, n = 33, S33=, , (a + l), , =, , (3 + 99), , =, , (102), , = 33 (51) = 1683, 12. Question, Find the sum of first n odd natural numbers., Answer, a = 1, d = 2, Sum of first n odd numbers, Sn =, =, , [2a + (n – 1) d], , [2(1) + (n – 1) 2], , = n [1 + n – 1], = n2, 13. Question, Find the sum of all odd numbers between, (i) 0 and 50, (ii) 100 and 200, Answer, (i) 0 and 50
Page 67 :
a = 1, l = 49, n = 25, S25=, , [a + l], , =, , [1 + 49], , =, , (50) = 625, , (ii) 100 and 200, a = 101, l = 199, n = 50, S50=, , [a + l], , = 25 (101 + 199), = 25 (300) = 7500, 14. Question, Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667., Answer, a = 3, l = 999, n =, Sn=, , = 167, , (a + l), , =, , (3 + 999), , =, , (1002), , = 167 (501) = 83667, 15. Question, Find the sum of all integers between 84 and 719, which are multiples of 5., Answer, To find: the sum of all integers between 84 and 719, which are multiples of 5., Solution: The smallest and largest digits between 84 and 719 which are divisible by 5 is 85 and, 715.So the sequence will be 85,90,........... 715. a = 85, d = 5, l = 715From the formula l=a+(n1)dThe number of terms are:, n=, , +1, , n=, , +1, , n=, , n= 127
Page 68 :
So there are 127 terms between 84 and 719.Now to calculate the sum of 127 terms use the formula:, , ⇒ S127=, , (a + l), , ⇒ S127=, , (85 + 715), , ⇒ S127=, , (800) = 50800So the sum of the terms between 84 and 719 is 50800., , 16. Question, Find the sum of all integers between 50 and 500, which are multiples of 7., Answer, a = 56, d = 7, l = 497, n=, , +1, , =, =, , +1, +1=, , = 64, S64=, , (a + l), , = 32 (56 + 497), = 32 (553) = 17696, 17. Question, Find the sum of all even integers between 101 and 999., Answer, a = 102, l = 998, n=, S449=, =, =, , = 449, (a + l), (102 + 998), = 246950, , 18. Question, Find the sum of all integers between 100 and 550, which are multiples of 9.
Page 69 :
Answer, a = 108, L = 549, D=9, 549 = 108 + (n - 1)9, 549 = 99 + 9n, 9n = 450, n = 50, Sn=, , [216 + 49 (9)], , = 16425, 19. Question, In an A.P. if the first term is 22, the common difference is – 4 and the sum to n terms is 64, find n., Answer, a = 22, d = -4, Sn = 64, 64 =, , (2a + (n – 1) d), , n [44 – (n – 1)4] = 128, n (44 – 4n + 4) = 128, 48n – 4n2 = 128, n2 – 12n + 32 = 0, n2 – 8n – 4n + 32 = 0, (n – 8) (n – 4) = 0, n = 8 or n = 4, 20. Question, In an A.P. If the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?, Answer, According to question,, a5= 30, a + 4d = 30
Page 70 :
a = 30 – 4d (i), a12= 65, a + 11d = 65, 30 – 4d + 11d = 65 [from (i)], 7d = 35, d=5, Put d in (i), a = 30 – 4(5), = 10, S20=, , [2(a) + (20 – 1) d], , = 10 [2(10) + 19(5)], = 10 {20 + 95}, = 1150, 21. Question, Find the sum of the first, (i) 11 terms of the A.P :, (ii) 13 terms of the A.P :, (iii) 51 terms of the A.P. whose second term is 2 and fourth term is 8., Answer, (i) 11 terms of the A.P :, a = 2, d = 6 – 2 = 4, S11=, =, , [2(a) + 10d], [2(2) + 10(4)], , = 11 [2 + 20], = 242, (ii) 13 terms of the A.P :, a = -6, d = 0 + 6 = 6, S13=, , [2(a) + 12d], , = 13 [-6 + 6(6)]
Page 71 :
= 13 [-6 + 36], = 13 (30) = 390, (iii) 51 terms of the A.P. whose second term is 2 and fourth term is 8., a2= 2, a + d = 2 (i), a4 = 8, a + 3d = 8, 2 – d + 3d = 8, 2 + 2d = 8, d=3, a = -1, S51=, , [2(a) + 50d], , = 51 [-1 + 25(3)], = 51 (74), = 3774, 22. Question, Find the sum of, (i) The first 15 multiples of 8, (ii) The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6., (iii) All 3 – digit natural numbers which are divisible by 13., (iv) All 3 – digit natural numbers, which are multiples of 11., Answer, (i) The first 15 multiples of 8, a = 8, d = 8, S15=, , [2(8) + 14(8)], , = 15[8 + 56]= 15[64], = 960, (ii) The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6., (a) a = 3, d = 3
Page 72 :
S40=, , [2(a) + 39(d)], , = 20 [2(3) + 39(3)], = 60 (41) = 2460, (b) a = 6, d = 5, S40=, , [2(5) + 39(5)], , = 100 [41], = 4100, (iii) All 3 – digit natural numbers which are divisible by 13., a = 6, d = 6, S40=, , [2(6) + 39(6)], , = 120 (41) = 4920, (iv) All 3 – digit natural numbers, which are multiples of 11., a = 110, d = 11, l = 990, n=, , +1, , =, , +1, , =, , =, , = 81, S81=, , [2(110) + 80(11)], , = 8910 [1 + 4], = 44550, 23. Question, Find the sum :, (i), (ii), (iii), (iv), (v)
Page 73 :
(vi), (vii), (viii), Answer, (i), a = 2, d = 4 – 2 = 2, l = 200, n=, , +1, , =, , + 1 = 100, , S100=, , [2(2) + 99(2)], , = 100 [101] = 10100, (ii), a = 3, d = 11 – 3 = 8, l = 803, Sn=, , +1=, , +1, , = 101, S101=, , [2(3) + 100(8)], , = 101 (403) = 40703, (iii), a = -5, d = -8 + 5 = -3, l = -230, n=, =, , +1, = 76, , Sn= 38(5) [-2 – 45], = 190 (-47) = -8930, (iv), a = 1, d = 3 – 1 = 2, l = 199, n=, = 100, , +1
Page 74 :
S100=, , [2(1) + 99(2)], , = 100 (100) = 10000, (v), a = 7, d =, , -7=, , Last term, an = 84, an = a + (n – 1)d, 84 = 7 + (n – 1), 84 =, 84 * 2 = 7 + 7n, 168 = 7 + 7n, 7n = 168 – 7, 7n = 161, n = 23, Sum of nth term, Sn =, S23 =, , [2a + (n – 1)d], , [2(7) + (23 – 1) ], , =, , [14 + 22 * ], , =, , [14 + 77], , =, , * 91 =, , Hence, sum of given A.P. is, (vi), a = 34, d = 32 – 34 = -2, Last term, an = 10, an = a + (n – 1)d, 10 = 34 + (n – 1) (-2), 10 = 34 – 2n + 2, 2n = 34 – 10 + 2
Page 75 :
2n = 26, n = 13, Sum of n terms,, Sn =, , [2a + (n – 1) d], , S13 =, , [2(34) + (13 – 1) (-2)], , =, , [68 + 12(-2)], , =, , [68 – 24], , =, , * 44 = 286, , Hence, Sum of given A.P. is 286, (vii), a = 25, d = 3, Last term, an = 100, An = a + (n -1) d, 100 = 25 (n – 1)3, 75 = 3n – 3, 78 = 3n, n = 26, Sum of n terms, Sn =, =, , [2a + (n – 1) d], , [2(25) + (26 – 1) 3], , = 13 [50 + 25 * 3], = 13 * 125 = 1625, Hence, Sum of given A.P. is 1625, (viii), a = 18, d =, , - 18 =, , Last term, an =, an = a+ (n – 1)d
Page 76 :
= 18 + (n – 1) (, , ), , -99 = 36 + 5 – 5n, -140 = -5n, n = 28, S28=, , [a + l], , = 14 (18 -, , ), , = 7 (36 – 99), = 7 (-63) = -441, 24. Question, Find the sum of the first 15 terms of each of the following sequences having nth term as, (i), (ii), (iii), (iv), Answer, (i), Put n = 1, a1 = 3 + 4(1) = 7, Put n = 15, a15 = 3 + 4(15) = 63 = l, Sum of 15 terms, S15 =, =, , [7 + 63] = 525, , (ii), Put n = 1, b1 = 5 + 2(1) = 7, Put n = 15, b15 = 5 + 2(15) = 35 = l, , [a + l]
Page 77 :
Sum of 15 terms, S15=, =, , [a + l], , [7 + 35] = 315, , (iii), Put n = 1, x1 = 6 – 1 = 5, Put n = 15, x15 = 6 – 15 = -9, Sum of 15 terms, S15=, =, , [a + l], , [5 - 9] = -30, , (iv), Put n = 1, y1= 9 – 5(1) = 4, Put n = 15, y15= 9 – 5(15) = -66, Sum of 15 terms, S15=, =, , [a + l], , [4 - 66] = -465, , 25. Question, Find the sum of first 20 terms of these sequence whose nth term is, Answer, Given, an= An + B, Put n = 1, a1= A + B, Put n = 20, a20= 20A + B, S20=, , [a + l], , = 10 [A + B + 20A + B], = 10 [21A + 2B], = 210A + 20B, 26. Question, , .
Page 78 :
Find the sum of first 25 terms of an A.P. whose nth term is given by, , ., , Answer, Given, nth term, an = 2 – 3n, Put n = 1, a1 = 2 – 3(1) = -1, Put n = 25, a15= 2 – 3(15) = -43, Therefore, S25 =, =, , (-1 – 43), , (-44) = -925, , 27. Question, Find the sum of the first 25 terms of an A.P. whose nth term is given by, , ., , Answer, Given, an= 7 - 3n, Put n = 1, a1= 7 – 3(1) = 4, Put n = 25, a25= 7 – 3(25) = -68 = l, S25=, , [4 – 68], , = 25 [-32] = -800, 28. Question, Find the sum of the first 25 terms of an A.P. whose second and third terms are 14 and 18, respectively., Answer, a2= 14, a + d = 14 (i), a3= 18, a + 2d = 18 (ii), Subtracting (i) from (ii), we get, d=4, Putting the value of d in (i), we get, a = 14 – 4 = 10
Page 79 :
S25=, , [2(a) + 24(d)], , = 25 [58], = 1450, 29. Question, If the sum of 7 terms of an A.P. is 49 and that of 17 term is 289, find the sum of n terms., Answer, S7= 49, [2a + 6d] = 49, a + 3d = 7 (i), S17= 289, [2a + 16d] = 289, a + 8d = 17 (ii), Subtract (i) from (ii), we get, 5d = 10, d=2, Put d = 2 in (i), we get, a=7–6=1, Sn=, , [2(1) + (n – 1)2], , = n [1 + n – 1], = n2, 30. Question, The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and, the common difference., Answer, a = 5, l = 45, Sn= 400, [5 + 45] = 400, [50] = 400, n = 16
Page 80 :
16th term is 45, a16= 45, 5 + 15d = 45, 15d = 40, d=, 31. Question, In an A.P., the sum of first n terms is, , . Find its 25th term., , Answer, Given,, , Let an, an=, , be the nth term of the A.P., , Sn – Sn – 1, , = 3n2/2 + 13n/2 – 3(n – 1)2/2 – 13(n – 1)/2, =, , {n2 – (n – 1)2} +, , {n – (n – 1)}, , = 3n - +, = 3n +, , = 3n + 5, , Put n = 25, a25= 3(25) + 5, = 75 + 5 = 80, Therefore, 25th term is a25 = 80, 32. Question, Let there be an A.P. with first term ‘a’, common difference, d. If, of first n terms, find., (i), (ii), (iii), (iv), (v), , denotes its nth term and, , the sum
Page 81 :
(vi), Answer, (i), Given, an= 50, a + (n – 1) d = 50, 5 + (n – 1) 3 = 50, (n – 1)3 = 45, (n -1) = 15, n = 16, Sn=, , [a + l], , = 8 [5 + 50] = 8 [55] = 440, (ii), an= 4, d = 2, Sn= -14, a + (n – 1) 2 = 4, a + 2n = 6, and, [2a + (n – 1)2] = -14, n [ 2a + 2n – 2] = -14, or, [a + an] = -14, [a + 4] = -14, n [6 -2n +4] = -28, n [10 -2n] = -28, 2n2 – 10n – 28 = 0, 2(n2 – 5n – 14) = 0, (n + 2) (n – 7) = 0, n = -2, n = 7, Therefore, n = -2 is not a natural number. So, n = 7, (iii), Given, a = 3, n = 8, Sn= 192
Page 82 :
Sn =, , {2a + (n – 1) d], , 192 * 2 = 8 [6 + (8 – 1)d], = 6 + 7d, 48 = 6 + 7d, 7d = 42, d=6, (iv), an= 28, Sn = 144, n = 9, Sn=, , [a + l], , 144 =, , [a + 28], , = a = 28, a + 28 = 32, a=4, (v), Given, a = 8, an= 62 and Sn= 210, Sn= [a + l], 210 =, , [8 + 62], , 210 * 2 = n (70), n=, , =6, , a + (n – 1) d = 62, 8 + (6 – 1) d = 62, 5d = 54, d = 10.8, (vi), Given, a = 2, d = 8 and Sn = 90, 90 =, , [4 + (n – 1)d] {Therefore, Sn=, , 180 = n [4 + 8n – 8], , [2a + (n – 1)d]}
Page 83 :
8n2 – 4n – 180 = 0, 4 (2n2 – n – 45) = 0, 2n2 – n – 45 = 0, (2n + 1) (n – 5) = 0, Therefore, n =, , is not a natural number, , n=5, an= a + (n – 1)d, = 2 + 4(8) = 32, 33. Question, Aman saved Rs. 16500 in ten years. In each year after the first he saved Rs.100 more than he did in, the preceding year. How much did he save in the first year?, Answer, Given,, Man saved in 10 years, S10= 16500, In each year after first he saved, d = 100, We know, sum of n terms, Sn=, , [2a + (n – 1) d], , For 10 years, S10=, , [2a + (10 – 1) 100], , 16500 = 5 [2a + 9 * 100], = 2a + 900, 3300 = 2a + 900, 2a = 3300 – 900, 2a = 2400, a = 1200, Hence, Rs. 1200 saved by him in first year, 34. Question, Aman saved Rs.32 during the first year, Rs.36 in the second year and in this way he increases his, savings by Rs.4 every year. Find in what time his saving will be Rs.200., Answer, Given, A man saved in first year, a = 32
Page 84 :
A man saved in second year, a2 = 36, Increase saving, d = 4, In n years his saving will be 200, Sn = 200, We know,, Sn =, , [2a + (n – 1) d], , 200 =, , [2(32) + (n – 1) 4], , 400 = n [64 + 4n – 4], 400 = n [60 + 4n], 400 = 4n [15 + n], 100 = 15n + n2, n2 + 15n – 100 = 0, n2 + 20n – 5n – 100 = 0, n (n + 20) – 5 (n + 20) = 0, (n – 5) (n + 20) = 0, Here, n – 5 = 0, n = 5, n + 20 = 0, n = -20, The term can never be negative. So, we consider n = 5, Hence, in 5 years his saving will be Rs. 200, 35. Question, Aman arrange stop ay off a debt of Rs.3600 by 40 annual installments which for man arithmetic, series. When 30 of the installments are paid, he dies leaving one - third of the debt unpaid, find the, value of the first installment., Answer, Given, Total amount of payable in 40 annual installments, S40 = 3600, After 30 installment he died and leaving, , of the debt unpaid, , Which means, total in 30 installment, S30=, S30=, , of the debt, , * 3600 = 2400, , We know sum of n terms, Sn =, , [2a + (n – 1) d]
Page 85 :
For 30 installments, S30 =, , [2a + (30 – 1) d], , 2400 = 15 [2a + 29d], 160 = 2a + 29d, 2a = 160 – 29d, a=, , (i), , For 40 installments, S40 =, 3600 =, , [2a + (n – 1) d], , [2a + (40 – 1) d], , 180 = 2a + 39d, 2a = 180 – 39d, a=, , (ii), , From (i) and (ii), we get, =, 160 – 29d = 180 – 39d, 39d – 29d = 180 – 160, 10d = 20, d=2, Putting the value of d in (i), we get, a=, =, , =, , = 51, , Hence, value of first installment is 51, 36. Question, There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from, the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well, and here turns to the well after watering each tree to get water for the next. Find the total distance, the gardener will cover in order to water all the trees., Answer, Total numbers of trees = 25, Distance between trees = 5, Total distance between well and first tree is = 10 m
Page 86 :
Now, gardener return back to well after watering each tree then,, Distance between well and second tree is = 10 + 10 + 5 = 25, Distance between well and third tree = 25 + 5 + 5 = 35, Distance between well and fourth tree = 35 + 5 + 5 = 45, Distance between well and last tree without return back = 10 + 24(5) = 10 + 120 = 130, So, common difference = 35 – 25 = 10, Now, Total distance covered by 24 trees,, We knew, Sn=, =, , [2a + (n – 1) d], , [2(25) + (24 – 1) 10], , = 12 [50 + 23 * 10], = 12 [50 + 230], = 12 * 280, = 3360, Now, distance covered by 25 trees = 10 + 3360 = 3370, Total distance covered by gardener with return back = 3370 + 130 = 3500 m, Hence, total distance will covered by gardener is 3500., 37. Question, A man is employed to count Rs.10710. He counts at there of Rs.180 per minute for half an hour. After, this, he counts at the rate of Rs.3 less every minute than the preceding minute. Find the time taken, by him to count the entire amount., Answer, Total amount for counting = Rs. 10710, In 1 min he counts =Rs. 180, In half an hour (30 min) he count = Rs.180×30 =Rs.5400, Amount left count after half an hour, Sn= Rs.10710 –Rs. 5400 = 5310, Now, in 31st min he count 3 less than preceding minute = Rs.180 – Rs.3 = Rs.177, In 32nd min he count 3 less than preceding minute = Rs. 177 –Rs. 3 =Rs. 174, Then, Arithmetic progression formed is 177, 174,…, Here, difference between minutes = 174 – 177 = -3, We know, sum of n terms, Sn= (n/2) [2a + (n – 1) d]
Page 87 :
5310 = (n/2) [2(177) + (n – 1) -3], 5310 × 2 = n [354 – 3n + 3], 10620 = 354n – 3n2 + 3n, 10620 = 357n – 3n2, ⇒ 3n2 – 357n + 10620 = 0, On taking 3 common from the complete equation, we get,, ⇒ 3 (n2 – 119n + 3540) = 0, ⇒ n2 – 119n + 3540 = 0, Now, from factoring by splitting the middle term method, we get,, ⇒ n2 – 60n – 59n + 3540 = 0, ⇒ n (n – 60) – 59 (n -60) = 0, ⇒ (n – 59) (n – 60) = 0, Therefore, either n = 59 or n = 60We will use 59 as these are minutesSo The total time to calculate, whole amount is 59 + 30 = 89 min, = 1 hr 29 min, 38. Question, A piece of equipment cost a certain factory Rs.600,000. If it depreciates in value, 15% the first,, 13.5% the next year, 12% the third year, and soon. What will be its value at the end of 10 years, all, percentages applying to the original cost?, Answer, Given: A piece of equipment cost a certain factory Rs.600,000. If it depreciates in value, 15% the, first, 13.5% the next year, 12% the third year, and soon., To find: The value at the end of 10 years if all percentages applying to the original cost., Solution:Cost of equipment = 6, 00, 000, In 1 year the value depreciate by 15%, ⇒ The value of the equipment after first year= 6, 00, 000 *, , = 90, 000, , In 2 year depreciate by 13.5%, ⇒ The value of the equipment after second year= 6, 00, 000 *, , = 81, 000, , In 3 year depreciate by 12%, ⇒ The value of the equipment after third year= 6, 00, 000 *, Now A.P. is 90000, 81000, 72000,…, , = 72, 000
Page 88 :
So, common difference = 81000 – 90000 = -9, 000, We have to find total depreciation for 10 years,Since the value of depreciation is constant in any, consecutive years i.e its value is -9000. So to find the depreciation after ten years we will use the, formula:Sn=, S10=, , [2a + (n – 1) d], , [2(90, 000) + (10 – 1) (-9000)], , S10= 5 [180000 + 9 * (-9000)], S10= 5 [180000 – 81000], S10= 5 [99000] = 495000, Hence, cost of equipment at the end of 10 years = original cost – depreciation, = 6, 00, 000 – 4, 95, 000 = Rs. 1, 05, 000, 39. Question, A sum of Rs.700 is to be used to gives even cash prizes to students of a school for their overall, academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each, prize., Answer, Amount of money = 700, Total number of prize = 7, Each prize is Rs. 20 less than its preceding prize, Let, Prize of first prize = a, Prize of second prize = a – 20, Prize of third prize = a – 20 – 20 = a – 40, Here, A.P. is a, a – 20, a – 40,…, So, common difference, d = a – 20 – a = -20, We know, Sn =, 700 =, , [2a + (n – 1) d], , [2a + (7 – 1) -20], , = 2a + 6 (-20), , 2a = 320, a = 160, So, value of first prize = 160
Page 89 :
Value of second prize, 160 – 20 = 140, Value of third prize, 140 – 20 = 120, Value of fourth prize, 120 – 20 = 100, Value of fifth prize, 100 – 20 =80, Value of sixth prize, 80 – 20 = 60, And value of seventh prize, 60 – 20 = 40, Hence, value of prizes, Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, Rs 40., 40. Question, In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the, common differences., Answer, Given, First term, a = 8, Nth term, an = 33 = l, And sum of n terms, Sn =, 123 =, , [a + l], , [8 + 33], , 123 * 2 = n * 41, n=, , =, , =6, , We know,, an= a + (n – 1) d, 33 = 8 + (6 – 1) d, 33 – 8 = 5d, 25 = 5d, d=5, Hence, number of terms n = 6 and common difference, d = 5, 41. Question, In an A.P., the first term is 22, nth term is -11 and the sum to first n terms is 66. Find n and d, the, common difference., Answer, Given, First term, a = 22, nth term, an= -11 = l
Page 90 :
and sum of n terms, Sn= 66, We know sum of n terms, Sn=, 66 =, , [a + l], , [22 – 11], , 66 * 2 = n * 11, n=, , = 12, , we know, an= a + (n – 1) d, -11 = 22 + (12 – 1) d, -11 – 22 = 11d, -33 = 11d, d = -3, Hence, number of terms, n = 12 and common difference, d = -3, 42. Question, If the sum of the first n terms of an A.P. is 4n – n2, which is the first term? What is the sum of first, two terms? What is the second term? Similarly, find the third, the tenth and the nth terms., Answer, Given, Sn = 4n – n2, Putting n = 1, 2, S1 = 4(1) – (1)2 = 3, S2 = 4(2) – (2)2 = 4, We know, an = Sn – Sn – 1, For first term, a1 = S1 – S0 = 3 – 0 = 3, For term a2 = S2 – S1 =4 – 3 = 1, Now fore term a3,, S2 = 4(2) – (2)2 = 8 – 4 = 4, S3 = 4(3) – (3)2 = 12 – 9 = 3, a3 = S3 – S2 = 3 – 4 = -1, 43. Question, The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how, many terms are there and what is their sum?
Page 91 :
Answer, Given,, First term, a1 = 17, Last term, an = 350 = l, And difference, d = 9, We know,, an = a + (n – 1)d, 350 = 17 + (n – 1) (9), 350 = 8 + 9n, 342 = 9n, n = 38, We know sum of n terms, Sn =, S38 =, , [a + l], , [17 + 350], , = 19 * 367 = 6973, Hence, number of terms, n = 38, Sum of n terms, Sn = 6973, 44. Question, In an A.P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common, difference of the A.P., Answer, Given, First term, a = 2, Last term, an = l = 29, And, Sum of terms, Sn =, 155 =, , [a + l], , [2 + 29], , 155 * 2 = n * 31, n=, , =, , = 10, , We know, an = a + (n – 1) d, 29 = 2 = (10 -1) d, 29 – 2 = 9d
Page 92 :
27 = 9d, d=3, Hence, common difference, d = 3, 45. Question, In an A.P., the sum of first ten terms is – 150 and the sum of its next ten terms is-550. Find the A.P., Answer, Given: Sum of first ten terms that is, S10 = -150 and Sum of next ten terms is -550., To find: The A.P., Solution: Let, first term = a, Last term = anWe know in an A.P,an = a + (n – 1) d, For n=10,, an = a + (10 -1) d, an = a + 9d, We know, Sum of n terms, Sn =, , [a + l], , ...... (1), , l and an means the last term., Put the value of l in (1),So,, S10 =, , [a + a + 9d], , Substitute the known values,, -150 = 5 (2a + 9d), -150 = 5 (2a + 9d)-150= 10a + 45d, 10a = -150 – 45d, a=, , ..... (2), , For next 10 terms, First term = a11, Last term = a20, Since an = a + (n – 1) d,, For 11th term,a11 = a + 10d, For 20th term,a20 = a +19d
Page 93 :
And S'n =, , [a + l], , ...... (3), , Let the sum of next 10 terms to be S'10 ,, Now a will be equal to a11 in this case and l will be equal to a + 19d, Put the value of a and l in the equation (3).S’10 =, Substitute the known values,, -550 = 5 [2a + 29d], -550 = 10a + 145d, 10a = -550 – 145d, a=, , .....(4), , From (2) and (4), we get, =, -150 – 45d = -550 – 145d, -45d + 145d = -550 + 150, 100d = -400, d = -4, Put the value of d in (2), we get, a=, a=, We know A.P is of the form,a1 , a2 ,.......... ,an, Where a1 is the first term of an A.P,, d is the common difference.andan = a+ ( n-1 )dSo,, a1 = 3, a2 = a + d = 3 + (-4) = -1, a3 = a + 2d = 3 + 2(-4) = -5, Conclusion : The A.P. is 3, -2, -5,…, and a = 3, d = -4, 46. Question, , [a + 10d + a + 19d]
Page 94 :
Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term., Answer, Given, a = 10, S14 = 1505, S14 =, , [2a + (n – 1) d], , 1505 =, , [2(10) + (14 – 1) d], , 1505 = 7 [20 + 13d], 215 = 20 + 13d, 13d = 215 – 20, 13d = 195, d = 15, Now, a25 = a + (n – 1) d, = 10 + (25 – 1) 15, = 10 + 24 * 15, = 10 + 360 = 370, Hence, value of 25th term is 370, 47. Question, The sum of first n terms of an A.P. is 5n2 + 3n . If its mth term is 168, find the value of m. Also, find, the 20th term of this A.P., Answer, Given that, Sn = 5n2 + 3n, Put n = 1, S1= T1= 5 + 3 = 8, Put n = 2, S2= 5(2)2 + 3 + 2 = 26, T2= S2 – S1 = 26 – 8 = 18, S3= 5(3)2 + 3 + 3 = 54, T3= S3 – S2 = 54 – 26
Page 95 :
= 28, Therefore, first term, a = 8 and common difference = 18 – 8 = 10, Tm = a + (m – 1) d, 168 = 8 + (m – 1) 10, 168 = 8 + 10m – 10, 170 = 10m, m = 17, T20 = 8 + (20 – 1) 10, = 8 + 19 * 10 = 198, 48. Question, The sum of first q terms of an A.P. is 63q – 3q2. If its pth term is-60, find the value of p. Also, find the, 11th term of this A.P., Answer, Given that: Sq= 63q – 3q2, Put q = 1, S1 = T1 = 63 – 3 = 60, Put q = 2, S2 = 63 * 2 – 3 * (2)2, = 126 – 12 = 114, T2 = S2 – S1 = 114 – 60 = 54, Put q = 3, S3 = 63 * 3 – 3(3)2, = 189 – 27 = 162, T3 = S3 – S2, = 162 – 114 = 48, Therefore, first term of this A.P. is 60 and Common difference is 54 - 60 = -6, Tp = a + (p – 1) d, -60 = 60 + (p – 1) (-6), -120 = (p – 1) (-6), (p – 1) = 20
Page 96 :
p = 21, Now 11th term of this A.P, T11 = 60 + (11 – 1) (-6), = 60 – 60 = 0, 49. Question, The sum of first m terms of an A.P. is 4 m2 - m. If its nth term is 107, find the value of n. Also, find, the 21st term of this A.P., Answer, Sm = 4m2 – m, Put m = 1, S1 = T1 = 4 – 1 = 1, Put m = 2, S2 = 4(2)2 – 2 = 14, T2 = S2 – S1 = 14 – 3 = 11, Put m = 3, S3 = 4(3)2 – 3 = 33, T3 = S3 – S2, = 33 – 14 = 19, The first term of given A.P. is 3 and common difference, d = 11 – 3 = 8, nth term of the given A.P. is 107, 107 = 3 + (n – 1)8, 104 = (n – 1)8, (n – 1) = 13, n = 14, the 21st term of the given A.P., T21 = 3 + (21 – 1)8, = 3 + 160 = 163, 50. Question, The nth term of an A.P. is given by (-4n+15). Find the sum of first 20 terms of this A.P., Answer, Given: The nth term of an A.P. (-4n+15).
Page 97 :
To find: the sum of first 20 terms of this A.P., Solution:We have,, Tn = (-4n + 15), T1 = -4 + 15 = 11, T2 =( -4 × 2 )+ 15 = 7, T3 =( -4 × 3 ) + 15 = 3Hence the A.P is 11,7,3,........., The first term is 11 and the common difference is, d = T2 – T1 = 7 – 11 = -4, We calculate the sum of terms of A.P by using the formula:Sn=, , [2a + (n – 1) d], , Substitute the known values to get the sum., The sum of first 20 terms, S20 =, S20 =, , [(2 ×11) + (20 – 1) (-4)], , [(22) + (19) (-4)], , S20 = 10 (22 – 76) S20 = 10(-54)S20 = -540, 51. Question, Find the number of terms of the A.P. If 1 is added to each term of this A.P., then find the sum of all, terms of the A.P. thus obtained., Answer, The given A.P. is -12, -9, -6,…21, Here, a = -12, d = -9 – (-12) = 3, Let the number of terms be n, Tn = a + (n – 1) d, 21 = -12 + (n – 1) (3), 21 = -15 + 3n, 36 = 3n, n = 12, If 1 is added to each term than A.P. is, -11, -8, -5…20, Here, a = -11, d = 3, S12=, , [2(-11) + (12 -1) * 3], , = 6 (-22 + 33)
Page 98 :
= 66, 52. Question, The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P., Answer, We have sum of n terms, Sn = 3n2 + 4n, Put n = 1, S1 = T1 = 3(1)2 + 4 (1) = 7, Put n = 2, S2 = 3(2)2 + 4 (2) = 20, T2 = S2 – S1 = 20 – 7 = 13, Put n = 3, S3 = 3(3)2 + 4(3) = 39, T3 = S3 – S2 = 39 – 20 = 19, Therefore, first term is 7 and common difference, d = 13 – 7 = 6, The 25th term is, Tn= a + (n – 1) d, T25 = 7 + (25 – 1) * 6, = 7 + 24 * 6 = 151, 53. Question, In a school, students decide to plant trees in and around the school to reduce air pollution. It was, decided that the number of trees, that each section of each class will plant, will be double of the class, in which they are studying. If there are 1 to 12 classes in the school and each class has two sections,, find how many trees were planted by the students., Answer, Each class has to plant double the class they are studying, Therefore, first class will plant = 2, Second class will plant = 4, Same as, 12th class will plant = 24, Total number of plants to be planted = sum of all plants, =, =, , [a + l], [2 + 24]
Page 99 :
= 6 * 26 = 156, Therefore, there are 2 sections in each class, So, total trees = 2 * 156 = 312, 54. Question, The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1:5, find, the A.P., Answer, S7 =, , [2a + (7 – 1) d], , 182 =, , [2a + 6d], , 52 = 2a + 6d, a + 3d = 26 (i), Acc. To question,, a17 = 5a4, a + 16d = 5(a + 3d), d = 4a, d = 4(26 – 3d) [from (i)], 13d = 104, d=8, Putting the value of d in (i), we get, a = 26 – 3*8) = 2, a1 = a = 2, a2 =a + d = 2 + 8 = 10, a3 = a + 2d = 2 + 2(8) = 18, Hence, The A.P. is 2, 10, 18,…, 55. Question, The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P., Answer, Let ‘a’ be the first term and ‘d’ be the common difference, Sn= 3n2 + 6n, Firsdt term, a, S1 = 3(1)2 + 6(1)
Page 100 :
=3+6=9, S2 = 3(2)2 + 6(2), a + a + d = 12 + 12, 9 + 9 + d = 24, 18 + d = 24, d=6, Therefore, an = a + (n – 1) d, = 9 + (n – 1) 6, = 6n + 3, 56. Question, The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th, term of this A.P., Answer, S7 = 63, [2(a) + 6d] = 639, (i), , Hence, for next 7 terms first term will be the 8th term i.e. a + 7d, Sum of next 7 terms, S’7 =, , [2(a + 7d) + 6d], , 161 = 7 [a + 7d + 3d], 23 = a + 10d, 23 = 9 – 3d + 10d [From (i)], 14 = 7d, d=2, Putting the value of d in (i), we get, A = 9 – 3(2) = 3, Now, a28= a + 27d, = 3 + 27(2), = 3 + 54 = 57, 57. Question
Page 102 :
Answer, Sn = 4n2 + 2n, First term, a = 4(1)2 + 2(1), =4+2=6, Sum of first two terms, S2 = 4(2)2 + 2(2), a + a + d = 16 + 4, 6 + 6 + d = 20, d=8, Now the nth term, an = a + (n – 1) d, = 6 + (n – 1) 8, = 6 + 8n – 8, = 8n - 2, 60. Question, If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its nth term., Answer, a10 = a + 9d, a + 9d = 21 (i), S10 =, , [2a + 9d], , 120 = 5 [2a + 9d], 24 = 2a + 9d, 24 = 2 (21 – 9d) + 9d [From (i)], -18 = -9d, d=2, Putting the value of d in (i), we get, a = 21 – 9(2), =3, The nth term, an = a + (n – 1) d, = 3 + (n – 1) 2, = 3n + 2n – 2
Page 103 :
= 2n + 1, 61. Question, Ram kali would need Rs.1800 for admission fee and books etc., for her daughter to start going to, school from next year. She saved Rs.50 in the first month of this year and increased her monthly, saving by Rs.20. After a year, how much money will she save? Will she be able to fulfill her dream of, sending her daughter to school?, [CBSE2005,2014], Answer, a = 50, d = 20, an = a + (n – 1) d = l, a12 = 50 + 11 * 20 = 270, Sn =, =, , [a + l], [50 + 270], , = 12 * 160 = 1920, Yes, she will be able to fulfill her dream of sending her daughter to school., 62. Question, The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find, its common difference., Answer, First term, a = 5, Last term, l = 45, Sn = 400, [a + l] = 400, , n = 16, Now, n =, 16 =, , +1, +1, , 16d = 40 + d, 15d = 40, d=
Page 105 :
= 6 [4a + 14d – 2a – 3d], = 6 [2a + 11d], Since, L.H.S = R.H.S, Hence, proved, 65. Question, If the sum of first n terms of an A.P. is (3n2 + 7n), then find its nth term. Hence write its 20th term., Answer, Sn =, , (3n2 + 7n), , First term, a = S1 =, =, , [3(1)2 + 7(1)], , [3 + 7] = 5, , S2 =, , [3(2)2 + 7(2)], , a+a+d=, , [12 + 14], , 5 + 5 + d = 13, d=3, an = a + (n – 1)d, = 5 + (n – 1) 3, = 5 + 3n – 3, = 3n + 2, a20 = 3(20) + 2, = 60 + 2 = 62, Hence, it’s 20th term is 62, 66. Question, The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1:2. Find the, first and 15th term of the A.P., Answer, Let a be the first term and d be the common difference, Now, a6 / a13 =, =
Page 106 :
2a + 10d = a + 12d, a = 2d (i), Now sum of first n terms of A.P, Sn =, , [2a + (n – 1) d], , S9 =, , [2a + 8d], , 162 = 9 (a + 4d), a + 4d = 18, 2d + 4d = 18 [Using (i)], 6d = 18, d =3, Now from (i), we get, a=2*3=6, So, first term, a1 = a = 6, 15th term, a15 = a + 14d = 6 + 14(3), = 6 + 42 = 48, Therefore, First term is 6 and 15th term is 48, , CCE - Formative Assessment, 1. Question, Define an arithmetic progression., Answer, An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed, number to the preceding term except the first term. This fixed number is called the common, difference of the AP., 2. Question, Write the common difference of an A.P. whose nth term is an = 3n + 7., Answer, If an = 3n + 7, Then a1 = 3 (1) + 7 = 10, a2 = 3(2) + 7 = 13, a3 = 3(3) + 7 = 16
Page 107 :
d= common difference = an – a n–1, = a3 – a2, = 16 – 13 = 3, 3. Question, Which tem of the sequence 114, 109, 104, .... is the first negative term?, Answer, Here a = 114, d is common difference, d = a3 – a2 = a2 – a1 = –5, For finding first negative term, Tn < 0, a + (n – 1) d < 0, 114 + (n – 1) (– 5) < 0, 114 – 5n + 5 < 0, 119 – 5n < 0, – 5n < – 119, 5n > 119, n > 119/5, n > 23.8, Therefore first negative term is 24th term, 4. Question, Write the value of a30 – a10 for the A.P. 4, 9, 14, 19, ………., Answer, Here a1 = 4, a2 = 9, a3 = 14, d = common difference = a3 – a2 = a2 – a1 = 14 – 9 = 9 – 4 = 5, a30 = a + (n–1) d, a30 = 4 + (30 –1) 5, a30= 4 + 29 X 5, a30= 149
Page 108 :
Similarly, a10 = 4 + 9x5, a10= 49, a30 – a10 = 149 – 49 = 100, 5. Question, Write 5th term from the end of the A.P. 3, 5, 7, 9, ...., 201., Answer, Here a = 201, a2 = 5, a3 = 7, d = a3 – a2 = a2 – a1, =7–5=5–3=2, tn = a + (n –1)d, tn = 3 + (n–1)2 = 201, (n –1) 2 = 201 – 3 = 198, n–1=, , = 99, , n= 99 + 1 =100, 5th term from end = 96th term, T95 = 3 + (96 –1)2, T95 = 3 + 95 x 2, T95 = 3 + 190 = 193, 6. Question, Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P., Answer, Terms are in A.P. if common difference (d) is same between two continuous numbers., x + 10 – 2x = 3x + 2 – (x + 10), 10 – x = 2x – 8, 18 = 3x, x= 6, So the terms in A.P. are: 2(6), 6 + 10, 3(6) + 2, = 12, 16, 20, 7. Question
Page 109 :
Write the nth term of an A.P. the sum of whose n terms is Sn., Answer, First term = a, Sum up to first term = a, Last term (nth term) = an, Sum up to n terms = Sn, Second last term = an–1, Sum up to (n–1) th term = sn–1, Therefore, an = Sn–Sn–1, 8. Question, Write the sum of first n odd natural numbers., Answer, First n odd numbers are 1, 3, 5, 7, 9........., (2n–1) which forms an A.P, a= 1, l= 2n – 1 where l be the last term, Sn =, , (2a + (n–1) d) or, , =, , (1 + 2n –1), , =, , (a + l), , (2n), , =n (n), =n2, 9. Question, Write the sum of first n even natural numbers., Answer, First n even Natural numbers are 2, 4, 6,……, 2n, This forms an A.P where, a= 2, l= 2n where l is last term of the A.P, Sn = 𝑛/2 (2a + (n–1) d) or, =, , (2 + 2n), , (a + l)
Page 110 :
=, , [2(1 + n)], , =n (n + 1), 10. Question, 1f the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference., Answer, Let a1, a2, a3…….. an be the given A.P, Given, sum of n terms = 3n2 + 5n, Sn=3n2 + 5n……….(1), Putting n =1 in (1), Sn= 3x12 + 5x1, =3+5=8, Sum of first 1 terms = first term, first term = a = S1 = 8, Sn = 3n2 + 5n, Putting n= 2 in….. (1), S2= 3x 22 + 5x2, S2= 22, Sum of first two terms = first term + second term, S2= a1 + a2, S2 – a1 = a2, a2= 22 – 8 = 14, Thus a1= 8, a2 = 14, d= common difference = 14 – 8= 6, 11. Question, Write the expression for the common difference of an A.P. whose first term is a and nth term is b., Answer, The nth term of the A.P whose first term a1 and common difference is d is given by, an = a + (n–1) d, Here an is given as b
Page 111 :
so b = a + (n–1) d, =d, ⇒ d=, 12. Question, The first term of an A.P. is p and its common difference is q. Find its 10th term., Answer, Given: The first term of an A.P. is p and its common difference is q., To find: its 10th term., Solution:The first term of A.P is p.Common difference = q, We know an = a + (n–1)dWhere a is first term and d is common difference., So,a10 = p + (10-1)q, a10 = p + 9q, 13. Question, For what values of p are 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P.?, Answer, Given: 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P., To find: The value of p, Solution:Consider a1 = 2p + 1 , a2 = 13 and a3 = 5p – 3The A.P will be of the form 2p + 1, 13, 5p –, 3,.........Since the terms are in A.P so the common differences in them is same.⇒ a2 - a1 = a3 - a2, ⇒ 13– (2p + 1) = 5p –3 –13⇒ 13– 2p - 1 = 5p –3 –13, ⇒12– 2p = 5p –16, ⇒12+16 = 5p +2p, ⇒ 28 = 7p⇒ p= 4, 14. Question, If, , , a, 2 are three consecutive terms of an A.P., then find the value of a., , Answer, , , a, 2 are three terms in an A.P so there common difference will be same
Page 113 :
a13 = a + 12d = 64…………(2), Subtracting (1) from (2), 6d = 30, D=5, Multiplying (1) by 2, 2a + 12d = 68………………….(3), Subtracting (2) from (3), a=4, a18 = a + (n–1) d, a18 = 4 + (17) 5, a18 = 89 = C, 2. Question, If the sum of P terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be, A. 0, B. p – q, C. p + q, D. – (p + q), Answer, Let a be the first term and d is common difference of the A.P, then sum of n terms in A.P is Sn = ( )[ 2a + (n – 1) d], Here, sp= q, Sq =p, Sp = ( ) [2a + (p – 1) d], q=, , [2a + (p – 1) d], , = [2a + (p – 1) d] --------(1), Sq = p = (, , [2a + (q – 1) d], , = [2a + (q – 1) d] …………..(2)
Page 114 :
Subtract (1) from (2) we get, , (q – p) d =, , / pq -----------(3), , d = -2(q + p) / pq -----------(3), Sum of first (p + q) terms, , [from (1) and (3)], , 3. Question, If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is, A. 2, B. 3, C. 1, D. 4, Answer, Here, Sn= 3n2 + n, d=6, Putting n= 1, S1 = 3 + 1 = 4, Sum of first 1 term = first term = 4, 4. Question
Page 115 :
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of, terms will be, A. 5, B. 6, C. 7D. 8, Answer, a1 = 1 (first term), a11 = 11 = l (last term), Sn= 36, But, Sn =, 36=, , (a + l), , (12), , n= 6, 5. Question, If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164?, A. 26th, B. 27th, C. 28th, D. none of these, Answer, Here, Sn = 3n2 + 5n, S1 = a1 = 3 + 5 = 8, S2 = a1 + a2 = 12 + 10 = 22, ⇒ a2 = S2 – S1 = 22 – 8 = 14, S3 = a1 + a2 + a3 = 27 + 15 = 42, ⇒a3 = S3 – S2 = 42 – 22 = 20, ∴ Given AP is 8, 14, 20, ....., Thus a = 8, d = 6, Given tm = 164., 164 = [a + (n –1)d]
Page 116 :
164 = [(8) + (m –1)6], 164 = [8 + 6m – 6], 164 = [2 + 6m], 162 = 6m, m = 162 / 6., ∴ m = 27., 6. Question, If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is, A. 4n – 3, B. 3n – 4, C. 4n + 3, D. 3n + 4, Answer, Here Sn = 2n2 + 5n, Sum of the A.P with 1 term = S1 = 2 + 5 = 7 = first term, Sum of the A.P with 2 terms = 8 + 10 = 18, Sum of the A.P with 3 terms = 18 + 15 = 33, a2 = S2 – S1 = 18 – 7 = 11, d = a2 –a1 = 11 – 7 = 4, nth term = a + (n–1) d, = 7 + (n–1) 4, nth term = 4n + 3, 7. Question, If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third, of these terms is 273, then the third term is, A. 13, B. 9, C. 21, D. 17, Answer
Page 117 :
Let 3 consecutive terms A.P is a –d, a, a + d. and the sum is 51, So, (a –d) + a + (a + d) = 51, 3a –d + d = 51, 3a = 51, a = 17, The product of first and third terms = 273, So, (a –d) (a + d) = 273, a2 –d2 = 273, 172 –d2 = 273, 289 –d 2 = 273, d2 = 289 –273, d2 = 16, d=4, Third term = a + d = 17 + 4 = 21, 8. Question, If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least,, then the numbers are, A. 5, 10, 15, 20, B. 4, 10, 16, 22, C. 3, 7, 11, 15, D. none of these, Answer, Let the 4 numbers be a, a + d, a + 2d, a + 3d., Sum of 4 numbers in A.P = 50, a + a + d + a + 2d + a + 3d = 50, ⇒ 4a + 6d = 50, ⇒ 2a + 3d = 25 --------------(1), Given the greatest number is 4 times the least., 4(a) = a + 3d, 4a – a = 3d, a=d
Page 118 :
Putting, a = d in (1), we obtain, 5d = 25, d=5, ⇒a = 5, ∴ First four terms are 5, 10, 15, 20., 9. Question, Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given, by d = Sn – kSn–1 + Sn–2, then k =, A. 1, B. 2, C. 3, D. none of these., Answer, Let a be the first term, n be the number of terms and d be the common difference of AP., Given d = Sn – kSn–1 + Sn–2., Now let n = 3, So, AP is : a, a + d, a + 2d, And d = S3 – k S3–1 + S3–2, d = S3 – k S2 + S1 ..............(1), Sum of n terms of an AP is given as:, Sn = ( ) x {2a + (n–1) d}, Now S1 = a, S2 = ( ) x (2a + (2–1) d) (n =2), S2 = (2a + d), S3 = ( ) x (2a + (3–1)d) (n =3), S3 =, , x (2a + 2d), , S3 = 3(a + d), S3 = 3a + 3d, Putting values of S1, S2 and S3 in equation 1, we get
Page 119 :
d = 3a + 3d – k (2a + d) + a, d = 4a + 3d – k (2a + d), k (2a + d) = 4a + 3d – d, k (2a + d) = 4a + 2d, k (2a + d) = 2(2a + d), k=2, 10. Question, The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P., and the common difference is given by, , , then k=, , A. S, B. 2S, C. 3S, D. none of these, Answer, Let the common difference and number of terms of AP be d and n respectively., Last term of AP = an = l (given), l = a + (n–1) d, n=, S=, , + 1……………… (1), (2a + (n–1) d), , S=, , + 1) (2a + (, , S=, , + 1) (2a + l –a), , S=, , + 1) (a + l), , (, , + 1) =, =, =, , –1, , + 1 –1) d) …………… using (1)
Page 120 :
D=, , Comparing this with, , We get k =2S, 11. Question, If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers,, then k =, A., , B., , C., , D., Answer, Given: the sum of first n even natural numbers is equal to k times the sum of first n odd natural, numbers., To find: The value of k, Solution:Sum of terms of A.P = n/2 ( 2a + (n-1) d )First n even natural numbers are: 2, 4, 6, 8,, ............, It forms an AP where first term a = 2 and common difference d = 4 – 2 = 2, ⇒ sum of n even natural terms =, , x {2 x 2 + (n–1)2}=, , ( 4 + 2n - 2 )=, , ( 2 + 2n )= × 2 ( 1 + n, , ), = n (n + 1), First n odd natural numbers are: 1, 3, 5, 7, ............, It forms an AP where first term is a = 1 and the common difference d = 3 – 1 = 2, Now sum of n terms = {2 x 1 + (n–1)2}, =, , ( 2 + 2n - 2 )=, , =nxn, , (2n )
Page 121 :
= n2, Now, According to given condition, Sum of first n even numbers = k X (Sum of first n odd numbers), ⇒ n (n + 1) = k x n2, ⇒kxn=n+1, ⇒k=, 12. Question, If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is, A., , B., , C., , D. none of these, Answer, a, b and 2a are in A.P so a is the first term and 2a is the last term denoted by T and Tn respectively., Here Common difference = b – a, Tn = 2a = a + (n–1) (b–a), So n =, Sum =, =, , {first term + last term}, { 3a }, , =, 13. Question, If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms, of the series in odd places, then, , =
Page 122 :
A., , B., , C., , D., Answer, S1 =, , (2a + (n–1) d), , Out of these odd numbers of terms, there are, S2 =, , (2a + (, , terms in odd places, , –1) d), , Common difference of two odd places is 2d, S2 =, , (2a + (n–1) d), , Now,, =, , 14. Question, If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is, equal to, A., , B., , mn p, , C., , p3, , D., , (m + n) p2, , Answer
Page 123 :
Let first term = a and Common difference = d, ∴ According to the question, S n = n 2 p, Sn = n/2 (2a + (n–1) d) = n2p, 2a + (n–1) d = 2np……………….(1), And Sm = m2p, Sm= m/2 (2a + (m–1) d) = m2p, 2mp = (2a + (m–1) d)……………….(2), Subtracting 2 from 1, 2a + (n–1) d – 2a – (m–1) d = 2 np – 2 mp, d (n–1 –m + 1) = 2p (n– m), d = 2p, putting value of d in (1), 2a + (n–1) 2p = 2np, a + (n–1)p = np, a=p, now Sp = p/2 (2a + (p–1) d), putting value of a = p and d = 2p, Sp = p/2 (2p + (p–1) 2p), Sp = p3, 15. Question, If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to, A. 4, B. 6, C. 8, D. 10, Answer, We know that Sn =, Now it is given that, S2n = 3Sn
Page 124 :
=3, , ), , 2(2n + 1) = 3(n + 1), 4n + 2 = 3n + 3, n=1, now, S3n/Sn = (, , ), , =, =6, 16. Question, In an AP, Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp + q is equal to, A. 0, B. –(p + q), C. p + q, D. pq, Answer, Sp = {2A + (p–1) D} = q, given, Sq =, , (2A + (q–1) D} = p, given, , On Subtracting the second equation from 1st we get,, D = –2, Also On adding the two equations we get,, 2A +, , = p/q + q/p, , Now,, Sp + q =, =, , {2A + (p + q–1)D}, { + –, , =, , {2A + (p + q–1) D}, , + (p + q–1) D}
Page 125 :
[ By substituting the value of D ], = – (p + q), 17. Question, If Sr denotes the sum of the first r terms of an A.P. Then, S3n: (S2n — Sn) is, A. n, B. 3n, C. 3, D. none of these, Answer, S2n =, Sn =, , (2a + (2n–1) d), (2a + (n–1)d), , S3n =, , (2a + (n–1)d), , S2n –Sn =, =, , (2a + (2n–1) d) –, , (2a + (n–1)d), , (2a + (n –1) d), , S3n: (S2n — Sn) =, , (2a + (n–1)d) :, , (2a + (n –1) d), , =3, 18. Question, If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is, A. 3200, B. 1600, C. 200, D. 2800, Answer, Here a1= 2, D = common difference = 4, Sn = (2a + (n–1) d), S40 =, , (2 x 2 + (39) 4)
Page 126 :
= 20 (4 + 156), = 3,200, 19. Question, The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is, A. 5, B. 10, C. 12, D. 14, Answer, Here a = 3, a2= 7, a3 = 11, d= a3– a2 = a2– a1 = 4, Sn=, , (2a + (n–1) d), , 406 =, , (6 + (n–1)4), , 406 =, , (4n + 2), , 2n2 + n – 406 = 0, 2n2 + 29 n – 28 n – 406 = 0, n (2n + 29) – 14(2n + 29) = 0, (n – 14) (2n + 29) = 0, Number of terms cannot be negative and in fractions so n= 14, 20. Question, Sum of n terms of the series, , A., B. 2n (n + 1, C., , D. 1, , is
Page 127 :
Answer, The given A.P is, The simplified A.P is, , Here a=, d=, , =, , Sn =, , (2a + (n–1) d), , =, , (2, , + (n–1), , =, , /2 (1 + n), , ), , = n (n + 1)/, 21. Question, The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is, A. 501th, B. 502th, C. 508th, D. none of these, Answer, Here, a9 = 449, A449 = 9, Let a is the first term and d is the common difference of the AP, Given 9th term of AP = 499, a + 8d = 499 .....(1), Again 499th term of AP = 9, a + 498d = 9 .....(2), Now subtract equation 1 and 2, we get, a + 8d – (a + 498d) = 499 – 9, a + 8d – a – 498d = 499 – 9, –490d = 490
Page 128 :
d = –490/490, d = –1, Put value of d in equation1, we get, a – 8 = 499, a = 499 + 8, a = 507, Let nth term is equal is to zero, a + (n–1)d = 0, 507 – (n–1) = 0 (By putting value of a and d), 507 – n + 1 = 0, 508 – n = 0, n = 508, So 508th term of AP is zero., 22. Question, If, , are in A.P. Then, x =, , A. 5, B. 3, C. 1, D. 2, Answer, Here, , are in A.P, , so, , –1/, 1/2 = (x + 2)/(x + 5), x + 5= 2x + 4, x= 1, , = –2/
Page 129 :
23. Question, The nth term of an A.P., the sum of whose n terms is Sn, is, A. Sn + Sn–1, B. Sn – Sn–1, C. Sn + Sn + 1, D. Sn – Sn + 1, Answer, The sum of n terms of an A.P is given by Sn, Sn =, , (2a + (n–1) d), , If the sum of n terms is given that is Sn is given then the nth term is given by the formula, Tn = Sn– Sn–1, Where Sn–1 is sum of the (n–1)th term of the A.P., 24. Question, The common difference of an A.P., the sum of whose n terms is Sn, is, A. Sn – 2Sn–1 + Sn–2, B. Sn – 2Sn–1– Sn–2, C. Sn – Sn–2, D. Sn – Sn–1, Answer, an is the nth term of an A.P and a n–1 is the (n–1)th term of an A.P,, d = common difference, Sn = sum of n terms of an A.P, d= an – an–1, But an= Sn – Sn–1, And an–1= Sn–1 – Sn–2, So d= Sn – Sn–1 – (Sn–1 – Sn–2), d= Sn – 2 Sn–1 + Sn–2, 25. Question
Page 130 :
If the sums of n terms of two arithmetic progressions are in the ratio, , , then their nth terms, , are in the ratio, A., , B., , C., , D., Answer, The sums of n terms of two A.P`s are in ratio, Let a1 and d1 be the first term and common difference of the first A.P, respectively. Similarly let a2, and d2 be the common difference of the second A.P, respectively., According to the given condition, =, , =, , =, , Now the Nth term is given by a + (N–1) d, Equating the coefficients, , ⇒ n = 2N – 1, =, 26. Question, , =, , =
Page 131 :
If Sn, denote the sum of n terms of an A.P. with first term a and common difference d such that, is independent of x, then, A. d=a, B. d=2a, C. a = 2d, D. d = – a, Answer, Given AP in which First term = a, Common difference = d, Number of terms = n, And Sn denotes the sum of n terms, So, , and, Now,, , If d= 2a, Sx /Skx = (2×a) / (k2×2a), = 1/ k2, , ,
Page 132 :
So Sx /Skx is independent of, , if d = 2a, , 27. Question, If the first term of an A.P. is a and nth term is b, then its common difference is, A., , B., , C., , D., Answer, ‘a’ is the first term of the A.P,, ‘b’ is the (n)th term and ‘d’ is the common difference of the A.P., Then, we have b = a + (n–1) d, = a + (n + 1)d, Hence, ‘d’ =, 28. Question, The sum of first n odd natural numbers is, A. 2n – 1, B. 2n + 1, C. n2, D. n2 – 1, Answer, The sum of the first n odd numbers forms an arithmetic progression with first term equal to 1 and the, last term equal to 2(n−1) + 1, The formula for an arithmetic progression of n terms with first term a1 and last term an is, (a1 + an), Substituting these values, a1=1, an= 2(n−1) + 1
Page 133 :
⇒ n (1 + 2n−2 + 1)2 = n (2n) 2= n2, 29. Question, Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other, is 3. The difference between their 30th terms is, A. 11, B. 3, C. 8, D. 5, Answer, for 1st A.P a = 8 and d= common difference = d, The nth term of an A.P is given by the formula, an= a + (n–1) d, The required difference is, = 8 + (30 – 1) d – 3 – (30 –1) d, =5, 30. Question, If 18, a, b, –3 are in A.P., the a + b =, A. 19, B. 7, C. 11, D. 15, Answer, Given 18, a, b, –3 are in A.P, a– 18 = –3 – b, Because the common difference of the A.P. always remains the same, a + b = 15, 31. Question, The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is, A.
Page 134 :
B., , C., , D., Answer, Let a1 and a2 be the first terms of the two A.P. and d1 and d2 be the common difference of the two, A.P, According to the given condition,, , Substituting n= 35 in……… (1), , an = a + (n–1) d, , Or 179: 321, 32. Question, If, , A. 8, B. 7, C. 10, , , then n =
Page 136 :
D. 25th term, Answer, S = 3n2 + 5nS1 = a1 = 3 + 5 = 8S2 = a1 + a2 = 12 + 10 = 22 ⇒ a2 = S2 – S1 = 22 – 8 = 14S3 = a1, + a2 + a3 = 27 + 15 = 42⇒ a3 = S3 – S2 = 42 – 22 = 20∴ Given AP is 8,14,20,.....Thus a = 8, d =, 6Given tm = 164.164 = [a + (n –1)d]164 = [(8) + (m –1)6]164 = [8 + 6m – 6]164 = [2 + 6m]162=, 6mm = 162 / 6.∴ m = 27., 34. Question, If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is, A. n (n – 2), B. n (n + 2), C. n (n + 1), D. n (n –1), Answer, an= a + (n–1) d = 2n + 1 (given)…………………(1), Sn= n/2 (2a + (n–1) d), By putting n=1 in (1), a=3, similarly a2= 5, a3= 7, d= common difference = a2 – a1= 2, Sn = n/2 (6 + (n–1) 2), = n/2(2n + 4), = n (n + 2), 35. Question, If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio, A. 3: 2, B. 3: 1, C. 1 :3, D. 2 : 3, Answer, As 18th term : 11th term ratio is 3:2
Page 137 :
18th term is a + 17d and 11th term is a + 10d, , 3a + 30d = 2a + 34d, a=4d…………….(1), 21st term is a + 20d = 24d (putting value of a from (1)), And 5th term is a + 4d= 8d (putting value of a from (1)), Ratio of 21st term : 5th term is, = 3:1, 36. Question, The sum of first 20 odd natural numbers is, A. 100, B. 210, C. 400, D. 420, Answer, Here A.P is 1,3, 5, ……, a= 1, d= 2 and n = 20, Sn =, , (2a + (n–1) d), , Sn = 20/2 (2 x 1 + (19) 2), Sn = 10 (2 + 38), Sn= 400, 37. Question, The common difference of the A.P. is, , A. –1, B. 1, C. q, D. 2q, Answer, , , is
Page 138 :
Here A.P =, , d= a2–, , a1, , d, d=, d= –1, 38. Question, The common difference of the A.P., , is, , A., , B., C. –b, D. b, Answer, A.P =, d = common difference = a2–a1, =, d=, d= –b, 39. Question, The common difference of the A.P., A. 2b, B. –2b, C. 3, , is
Page 139 :
D. –3, Answer, Here A.P is, d= a2–a1, d=, d=, d= –3, 40. Question, If k, 2k –1 and 2k + 1 are three consecutive terms of an AP, the value of k is, A. – 2, B. 3, C. – 3, D. 6, Answer, Here A.P = k, 2k –1, 2k + 1, Since the numbers are in A.P their common difference (d) should be same, d=a2–a1 = a3–a2, 2k–1–k = 2k + 1– (2k –1), k– 1 = 2, k=3, 41. Question, The next term of the A.P., A., B., C., D., Answer
Page 140 :
A.P is, The simplified A.P is √7, 2√7 3√7, , d= 2√7 – √7 = √7 (2–1) = √7, Next term of A.P means the fourth term = a4, a4 = a + (n–1) d, a4= √7 + 3√7, = 4√7, = √112, , 42. Question, The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals, A. –3, B. 4, C. 5, D. 2, Answer, A.P here is 3y – 1, 3y + 5 and 5y + 1., So d= common difference = a2– a1= a3–a2, 3y + 5 – (3y–1) = (5y + 1) – (3y + 5), 6 = 2y – 4, 10 / 2 = y, Y=5
