Page 1 : Arithmetic Progression, Solved Problem Set - 1, 1., , The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term., , Solution:, a4= 0, a + 3d = 0, a = - 3d ------------- (eq. 1), It has to be proven, a25= 3 (a11), a + 24d = 3 (a + 10d), a + 24d = 3a + 30d, In LHS substitute a = - 3d ( from eq.1), - 3d + 24d = 21d, Now substitute a = - 3d in LHS ( from eq.1), 3a + 30d, - 9d + 30d = 21d, So, LHS = RHS, Hence proved, 2., , Find the 20th term from the end of the AP 3, 8, 13……..253., , Solution:, 3, 8, 13 ………….. 253, Last term = 253 and d = 5, a20 from end, = l – (n-1)d, =253 – ( 20-1) 5, =253 – 95, = 158 Ans, 3., , If the pth, qth & rth term of an AP is x, y and z respectively,, show that x(q-r) + y(r-p) + z(p-q) = 0, , Solution:, pth term x = a + (p-1) d, qth term y = a + (q-1) d, rth term z = a + (r-1) d, To prove x(q-r) + y(r-p) + z(p-q) = 0, {a+(p-1)d}(q-r) + {a + (q-1)d} (r-p) + {a+(r-1)d} (p-q), = a {(q-r) + (r-p) + (p-q)} + d {(p-1)(q-r) + (r-1) (r-p) + (r-1) (p-q)}, = a.0 + d{p(q-r) + q(r-p) + r (p-q) - (q-r) – (r-p)-(p-q)}, = A.0 + D.0 = 0., Hence proved, , Visit us on: www.k2pclasses.com Write us on:
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Page 2 : Arithmetic Progression, Solved Problem Set - 1, 4., , Find the sum of first 40 positive integers divisible by 6 also find the sum of first 20, positive integers divisible by 5 or 6., , Solution:, First 40 numbers divisible by 6 are, 6, 12 ……………. 240., 40, 6 240, S40 =, 2, = 20 x 246, = 4920 Ans, First 20 numbers divisible by 5 or 6 are (LCM of 6 and 5 is 30, so it means divisible by 30), 30, 60 …………. 600, 20, S20 = 30 600, 2, = 10 x 630, = 6300 Ans, 5., , A man arranges to pay a debt of Rs.3600 in 40 monthly installments which are in a AP., When 30 installments are paid he dies leaving one third of the debt unpaid. Find the, value of the first installment., , Solution:, Let the value of the first installment be a and c.d. be d, S40 = 3600., 40, 2a 39d =3600, 2, 2a + 39d = 180 …………………. (eq. 1), 30, 2a 29d =2400, S30 =, 2, 30a + 435d = 2400, 2a + 29d = 160 ………………… (eq. 2), Solving eq. 1 and eq. 2, we get, d = 2 and a = 51, Value of the first installment = Rs.51 Ans, 6., , Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5., , Solution:, We have to find out, 103+108+113+……….+998, Here a + (n-1)d = 998 (formula of nth term), 103 + (n-1)5 = 998, , n = 180, 180, 103 998= 90 x 1101=99090 Ans, So required sum =, 2, Visit us on: www.k2pclasses.com Write us on:
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Page 3 : Arithmetic Progression, Solved Problem Set - 1, 7., , Find the value of x if 2x + 1, x2 + x +1, 3x2 - 3x +3 are consecutive terms of an AP., , Solution:, As the given terms are in AP, difference between 2nd and 1st terms will be equal to difference, between 3rd and 2nd terms, So, a2 –a1 = a3 –a2, x2 + x + 1-2x - 1 = 3x2 – 3x + 3- x2-x-1, x2 - x = 2x2 – 4x + 2, x2 - 3x + 2 = 0, (x -1) (x – 2) = 0, x = 1 or x = 2 Ans, 8. Raghav buys a shop for Rs.1,20,000.He pays half the balance of the amount in cash and, agrees to pay the balance in 12 annual instalments of Rs.5000 each. If the rate of interest, is 12% and he pays with the instalment the interest due for the unpaid amount. Find the, total cost of the shop., Solution:, Balance = Rs.60,000 in 12 instalment of Rs.5000 each., 12, 1st instalment = 5000 +, 60,000 = 12200, 100, 2nd instalment = 5000 + (Interest on unpaid amount), 12, , = 5000 + 6600, 100 x 55000, = 11600, rd, 3 instalment = 5000 + (Interest on unpaid amount of Rs.50,000) =11000, AP is 12200, 11600, 11000, Common difference is 600, Cost of shop = 60000 + [sum of 12 instalment], 12, = 60000 +, [2x12200-11x600], 2, = 1,66,800 Ans, 9. Prove that am + n + am - n =2am, Solution:, a m+n = a1 + (m + n - 1) d ………….. (eq.1), a m-n = a1 + (m - n -1) d …………..(eq. 2), am = a1 + (m - 1) d, Add eq.1 & eq. 2, a m+n + a m-n = a1+(m+n-1) d+ a1 + (m-n-1)d, = 2a1+(m+n+m-n-1-1)d, = 2a1+ 2(m-1)d, = 2[a1+ (m-1)d], = 2[a1+ (m-1)d], = 2am. Hence proved, , Visit us on: www.k2pclasses.com Write us on:
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Page 4 : Arithmetic Progression, Solved Problem Set - 1, 10. If the roots of the equation (b-c)x2 +(c-a)x +(a-b) = 0 are equal show that a, b, c are in AP., Solution:, If (b-c)x2 + (c-a)x + (a-b) have equal root., B2-4AC = 0, (c-a)2-4(b-c)(a-b) = 0 (simplify and you will get), b-a=c-b, a, b, c are in AP Hence Proved, 11. Balls are arranged in rows to form an equilateral triangle .The first row consists of one, ball, the second two balls and so on. If 669 more balls are added, then all the balls can be, arranged in the shape of a square and each of its sides then contains 8 balls less than each, side of the triangle. Find the initial number of balls., Solution:, Let there be n balls in each side of the triangle, No. of ball (in ) = 1 + 2 + 3……….. + n =, , nn 1, 2, , No. of balls in each side square = n-8, No. of balls in square = (n-8)2, nn 1, APQ, + 660 = (n-8)2, 2, On solving, n2 + n + 1320 = 2(n2 - 16n + 64), n2 - 33n - 1210 = 0, (n-55) (n+22) = 0, n = -22 (Not Possible), n = 55, nn 1 55x 56, No. of balls =, =, 2, 2, = 1540 Ans, 1, 2, 3, 12. Find the sum of (1 ) (1 ) (1 ) +……. up to n terms., n, n, n, Solution:, 1, 2, 3, (1 ) (1 ) (1 ) +……. up to n terms, n, n, n, 1 2, = [1+1+…….+n terms] – [ + +….+ n terms], n n, 1, = n – [1 + 2 + 3 + ....up to n terms], n, 1 n(n 1), =n2, n, n 1 n 1, So, sum = n =, Ans, 2, 2, , Visit us on: www.k2pclasses.com Write us on:
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Page 5 : Arithmetic Progression, Solved Problem Set - 1, , 13. If the following terms form an AP. Find the common difference & write the next 3 terms, 3, 3+ 2, 3+22, 3+32………., Solution:, , d=, , 2 next three terms 3 + 4 2 , 3 + 5 2 , 3 + 6 2 …….. Ans, , 14. Find the sum of a+b, a-b, a-3b, …… to 22 terms., Solution:, a + b, a – b, a – 3b, up to 22 terms, d= a – b – a – b = -2b, 22, S22 = [2(a+b)+21(-2b)], 2, 11[2a + 2b – 42b], = 22a – 440b Ans., 15. Write the next two terms 12, 27, 48, 75………………., Solution: Next two terms 108 , 147 AP is 2 3 , 3 3 , 4 3 , 5 3 , 6 3 , 7 3 ……Ans, 16. If the pth term of an AP is q and the qth term is p. Prove that its nth term is (p+q-n)., Solution:, Let a be the first term and d be the common difference., tp = q, tq = p, tn = ?, a + (p-1) d = q (given) …….. (eq. 1), a + (q-1) d = p (given) ……...(eq. 2), Operating eq.1 – eq.2, d[p – q] = q – p, d = -1, Substituting d = -1 in eq. 1 we get a = q + p -1, an = a + (n – 1)d, = a + (n - 1)d, = (q + p – 1) + (n – 1) (- 1) = q + p – 1 – n + 1, = (q + p – n) Hence Proved, 1, 1, 1, ,, ,, 17. If, are in AP find x., x2 x3 x5, Solution : Since the three terms are in AP so, 1, 1, 1, 1, , , , x3 x2 x5 x3, 1, 2, 2, 2, x 5 x 6 x 8 x 15, On solving we get x = 1 Ans, Visit us on: www.k2pclasses.com Write us on:
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Page 6 : Arithmetic Progression, Solved Problem Set - 1, , 18. Find the middle term of the AP 1, 8, 15….505., Solution: First, We will find number of terms n first, a + (n-1)d = 505, a + (n-1)7 = 505, 504, n–1=, 7, n = 73, 37th term is middle term, a37 = a + 36d, = 1 + 36(7), = 1 + 252, = 253 Ans, 19. Find the common difference of an AP whose first term is 100 and sum of whose first 6, terms is 5 times the sum of next 6 terms., Solution:, Given a = 100, And a1 + a2 + ……. a6 = 5 (a7 + …….. + a12), a 1 a6 , a a12 , = 5 x 6 7, 6, , 2 , , 2 , , , , , a + a + 5d = 5[a + 6d + a + 11d], 8a + 80d = 0 (a = 100), d = - 10 Ans, , 20. Find the sum of all natural no. between 101 & 304 which are divisible by 3 or 5., Find their sum., Solution:, No. which are divisible by 3 are : 102, 105………….303 (68 terms), No. which are divisible by 5 are : 105, 110……300 (40 terms), No. which are divisible by 15 (3 & 5) are : 105, 120……300 (14 terms), There are 94 terms between 101 & 304 divisible by 3 or 5. (68 + 40 – 14), sum of required numbers = S68 + S40 – S14, So, sum of required numbers = 19035 Ans, , Visit us on: www.k2pclasses.com Write us on:
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Page 8 : Arithmetic Progression, Solved Problem Set - 1, 23. Find the sum of all natural numbers amongst first one thousand numbers which are, neither divisible 2 or by 5, Solution:, Sum of all natural numbers in first 1000 integers which are not divisible by 2, i.e. sum of odd integers., 1 + 3 + 5 + ………. + 999, Here, n = 500, 500, Hence, S500 =, [1 + 999], 2, = 250000, Sum of numbers which are divisible by 5 are, 5 + 15 + 25 …….. + 995, Here, n = 100, 100, Hence, S100 =, [5 + 995], 2, = 50 x 1000 = 50000, Required sum = 250000 – 50000, = 200000 Ans, , Visit us on: www.k2pclasses.com Write us on:
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