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ARITHMETIC, PROGRESSION, (A) OBJECTIVE TYPE QUESTIONS, Stand Alone MCQs, , (C) an A.P., with d = – 4, (1 Mark Each), , 1. 30th term of the A.P., : 10, 7, 4,………, is:, (A) 97, (B) 77, (C) –77, (D) –87, Ans. Option (C) is correct., , R, , (C) –38, , 1, (D) −48 , 2, , 5, 5, 6. The 11th term of the A.P.,: − 5 , − , 0 , , ... is:, 2, 2, R, , Explanation: In the given A.P.,, 1, 5, a = –3 and d = − + 3 =, 2, 2, , (A) –20, (C) –30, Ans. Option (B) is correct., , (B) 20, (D) 30, , R, , Explanation: In the given A.P.,, 5, 5, a = –5, d = − − ( −5 ) = , n = 11, 2, 2, , 5, Thus, the 11th term, t11 = −3 + (11 − 1)   = 22, 2, , 5, tn = a + ( n − 1) d ⇒ t11 = −5 + (11 − 1)   ⇒ t11 = 20, 2, R, , Explanation: In the given A.P.,, d = –4, n = 7, an = 4, an = a + ( n − 1) d ⇒ 4 = a + ( 7 − 1) ( −4 ) ⇒ a = 28, 4. In an A.P., if a = 3.5, d = 0, n = 101, then an will be, (A) 0, (B) 3.5, (C) 103.5, (D) 104.5, R, Ans. Option (B) is correct., Explanation: In the given A.P.,, a = 3.5, d = 0, n = 101, an = a + ( n − 1) d ⇒ an = 3.5 + (101 − 1) 0 ⇒ an = 3.5, 5. The list of numbers – 10, – 6, – 2, 2,…… is:, (A) an A.P., with d = – 16, (B) an A.P., with d = 4, , Explanation: In the given numbers, – 10, – 6, – 2, 2,…, ( −6 ) − ( −10 ) = 4, , thus, the given numbers are in A.P. with d = 4., , Ans. Option (B) is correct., , 3. In an A.P., if d = –4, n = 7, an = 4, then a is;, (A) 6, (B) 7, (C) 20, (D) 28, Ans. Option (D) is correct., , U, , Since, ( −6 ) − ( −10 ) = ( −2 ) − ( −6 ) = 2 − ( −2 ) = 4 ,, , 1, 2. 11th term of the A.P., : − 3 , − , 2 , ... is:, 2, (B) 22, , (D) not an A.P.,, Ans. Option (B) is correct., , ( −2 ) − ( −6 ) = 4, 2 − ( −2 ) = 4, , Explanation: In the given A.P.,, a = 10 and d = 7 – 10 = –3, Thus, the 30th term, t30 = 10 + (30 – 1) (–3) = –77, , (A) 28, , 1 Mark Each, , 7. The first four terms of an A.P., whose first term is, –2 and the common difference is –2, are:, (A) – 2, 0, 2, 4, (B) – 2, 4, – 8, 16, (C) – 2, – 4, – 6, – 8, Ans. Option (C) is correct., , (D) – 2, – 4, – 8, –16, , U, , Explanation: In the given A.P., a = –2, d = –2,, tn = a + ( n − 1) d, t1 = ( −2 ) + (1 − 1) ( −2 ) = −2, , t2 = ( −2 ) + ( 2 − 1) ( −2 ) = −4, t3 = ( −2 ) + ( 3 − 1) ( −2 ) = −6, , t4 = ( −2 ) + ( 4 − 1) ( −2 ) = −8, 8. The 21st term of the A.P., whose first two terms are, –3 and 4 is :, (A) 17, (B) 137, (C) 143, (D) –143, R, Ans. Option (B) is correct.

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Explanation: In the given A.P.,, t1 = –3 and t2 = 4, d = t2 − t1 = 4 − ( −3 ) = 7, ⇒, tn = a + ( n − 1 ) d, , ⇒, , t21 = ( −3 ) + ( 21 − 1)( 7 ) = 137, , 9. The famous mathematician associated with, finding the sum of the first 100 natural numbers is:, (A) Pythagoras, (B) Newton, (C) Gauss, (D) Euclid, R, Ans. Option (C) is correct., Explanation: The famous mathematician, associated with finding the sum of the first 100, natural numbers is Gauss., 10. If the first term of an A.P. is –5 and the common, difference is 2, then the sum of the first 6 terms is:, (A) 0, (B) 5, (C) 6, (D) 15, R, Ans. Option (A) is correct., Explanation: In the given A.P.,, a = –5 and d = 2, n, Sn =  2 a + ( n − 1) d , 2, 6, S6 =  2 × ( −5 ) + ( 6 − 1) × 2 , ⇒, 2, =0, 11. The sum of first 16 terms of the A.P.,: 10, 6, 2,... is:, (A) –320, (B) 320, (C) –352, (D) –400, R, Ans. Option (A) is correct., Explanation: In the given A.P.,, a = 10, d = 6 – 10 = –4, Thus,, n, Sn =  2 a + ( n − 1) d , 2, 16, ⇒ S16 =  2 × 10 + (16 − 1) × ( −4 ) , 2, = −320, 12. In an A.P., if a = 1, an = 20 and Sn = 399, then n is:, (A) 19, (B) 21, (C) 38, (D) 42, R, Ans. Option (C) is correct., Explanation: In the given A.P., a = 1, a n = 20, and S n = 399, an = a + ( n − 1 ) d, , ⇒, ⇒, , 20 = 1 + ( n − 1) d, , ( n − 1) d = 19, , n,  2 a + ( n − 1) d , 2, n, 399 = [ 2 + 19 ], 2, n = 38, Sn =, , ⇒, ⇒, , 13. The sum of first five multiples of 3 is:, (A) 45, (B) 55, (C) 65, Ans. Option (A) is correct., , (D) 75, , U, , Explanation: In the given A.P.,, a = 3 , d = 3 and n = 5, n, Sn =  2 a + ( n − 1) d , 2, 5, S5 =  2 × 3 + ( 5 − 1) × 3  = 45, ⇒, 2, 14. The sum of first five positive integers divisible by, 6 is:, (A) 180, (B) 90, (C) 45, (D) 30, R, Ans. Option (B) is correct., Explanation: Positive integers divisible by 6 are, 6, 12, 18, 24, 30, Since difference is same, its an A.P., We need to find sum of first 5 integers, We can use formula, n, Sn =, [2a + (n – 1) d], 2, Here,, \, , n = 5, d = 6, a = 6, 5, S5 =, [2 × 6 + (5 – 1) × 6], 2, S5 =, , 5, [12 + 24], 2, , 5, S5 =, × 36, 2, , = 90., , Case-based MCQs, , (1 Mark Each), , Attempt any four sub-parts from each question., Each sub-part carries 1 mark., I. Read the following text and answer the questions, given below it:, Your friend Veer wants to participate in a 200 m, race. He can currently run that distance in 51, seconds and with each day of practice it takes, him 2 seconds less. He wants to do in 31 seconds., , [CBSE QB, 2021]

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1. Which of the following terms are in A.P. for the, given situation, (A) 51, 53, 55…., , (B) 51, 49, 47…., , (C) – 51, – 53, – 55…., , (D) 51, 55, 59…, , Ans. Option (B) is correct., Explanation:, , a = 51, d =–2, A.P. = 51, 49, 47 ...... ., , II. Read the following text and answer the questions, given below it:, Your elder brother wants to buy a car and plans, to take loan from a bank for his car. He repays, his total loan of ` 1,18,000 by paying every month, starting with the first instalment of ` 1000. If he, increases the instalment by ` 100 every month., , [CBSE QB, 2021], , 2. What is the minimum number of days he needs to, practice till his goal is achieved ?, (A) 10, (C) 11, Ans. Option (C) is correct., Explanation:, , (B) 12, (D) 9, , Goal = 31 second, n = number of days, , \, , an = 31, a + (n – 1)d = 31, , 1. The amount paid by him in 30th instalment is:, , 51 + (n – 1)(– 2) = 31, , (A) ` 3900, , (B) ` 3500, , 51 – 2n + 2 = 31, , (C) ` 3700, , (D) ` 3600, , – 2n = 31 – 53, – 2n = – 22, n = 11, 3. Which of the following term is not in the A.P. for, the above given situation, (A) 41, , (B) 30, , (C) 37, , (D) 39, , Ans. Option (B) is correct., 4. If nth term of an A.P. is given by an = 2n + 3 then, common difference of an A.P. is:, (A) 2, , (B) 3, , (C) 5, , (D) 1, , Ans. Option (A) is correct., , Ans. Option (A) is correct., Explanation:, , a = 1000, d = 100, a30 = a + (30 – 1)d, , = 1000 + (30 – 1)100, = 1000 + 2900, = ` 3900, 2. The amount paid by him in the 30 instalments is:, (A) ` 37000, , (B) ` 73500, , (C) ` 75300, , (D) ` 75000, , Ans. Option (B) is correct., Explanation: Sum of 30 instalments, , Explanation:, Here,, , a1 = 2(1) + 3 = 5, a2 = 2(2) + 3 = 7, \, d = a2 – a1, = 7 – 5 = 2, 5. The value of x, for which 2x, x + 10, 3x + 2 are three, consecutive terms of an A.P., , =, , n, [2a + (n – 1)d], 2, , =, , 30, [2 × 1000 + (30 – 1)100], 2, , = 15[2000 + 2900], = 15 × 4900, , (A) 6, , (B) – 6, , = 73500, , (C) 18, , (D) – 18, , Total amount paid in 30 instalments = ` 73500, , Ans. Option (A) is correct., Explanation: Since, 2x, x + 10, 3x + 2 are in A.P.,, then common difference will remain same., x + 10 – 2x = (3x + 2) – (x + 10), 10 – x = 2x – 8, 3x = 18, x =6, , 3. What amount does he still have to pay after 30th, instalment ?, (A) ` 45500, (B) ` 49000, (C) ` 44500, (D) ` 54000, Ans. Option (C) is correct., Explanation:, Remaining amount = ` 1,18,000 – ` 73,500, = ` 44,500

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4. If total instalments are 40, then amount paid in the, last instalment ?, (A) ` 4900, (B) ` 3900, (C) ` 5900, (D) ` 9400, Ans. Option (A) is correct., Explanation: Amount paid in 40th instalment,, , a40 = 1000 + (40 – 1)100, = 1000 + 3900, = ` 4900, 5. The ratio of the 1st instalment to the last instalment, is:, (A) 1 : 49, (B) 10 : 49, (C) 10 : 39, (D) 39 : 10, Ans. Option (B) is correct., , Ans. Option (B) is correct., Explanation: Since the top and the bottom, 1, 5, rungs are apart by 2 m = m, 2, 2, 5, =, × 100 cm, 2, = 250 cm, 2. Total number of the rungs is:, (A) 20, (B) 25, (C) 11, (D) 15, Ans. Option (C) is correct., Explanation: The distance between the two, rungs is 25 cm., 250, Hence, the total number of rungs =, +1, 25, = 11., , Explanation: 1st instalment : last instalment, = 1000 : 4900, = 10 : 49, III. Read the following text and answer the questions, given below it:, A ladder has rungs 25 cm apart. (see the fig. below)., , 3. The given problem is based on A.P. find its first, term., (A) 25, (C) 11, Ans. Option (A) is correct., , (B) 45, (D) 13, , Explanation: The length of the rungs increases, from 25 to 45 and total number of rungs is 11., Thus, this is in the form of an A.P., whose first, term is 25., , 25 cm, , 4. What is the last term of A.P. ?, 2, , (A) 25, (C) 11, , 1, m, 2, , (B) 45, (D) 13, , Ans. Option (B) is correct., Explanation: Total number of terms, n = 11 and, the last term, T11 = 45., , 25 cm, 45 cm, , 5. What is the length of the wood required for the, rungs ?, (A) 385 cm, (B) 538 cm, (C) 532 cm, (D) 382 cm, Ans. Option (A) is correct., , The rungs decrease uniformly in length from, 45 cm at the bottom to 25 cm at the top. The top and, 1, the bottom rungs are 2 m apart., 2, , Explanation: The required length of the wood,, 11, S11 =, [25 + 45], 2, =, , 1. The top and bottom rungs are apart at a distance:, (A) 200 cm, (B) 250 cm, (C) 300 cm, (D) 150 cm, , 11, × 70, 2, , = 385 cm., , (B) SUBJECTIVE QUESTIONS, Very Short Answer Type Questions, (1 Mark Each), 1. Which term of the following A.P. 27, 24, 21, ........ is, zero ? , A [CBSE SQP, 2020-21], , Sol. We know that, , an = a + (n – 1)d, , Here,, , an = 0, , , , , 0 = 27 + (n – 1)(– 3), 30 = 3n, , ½

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n = 10, ½, 10th term of the given A.P. is zero., , [CBSE Marking Scheme, 2020-21], , , \Common difference = a2 – a1 =, , Detailed Solution:, Given A.P. = 27, 24, 21, ........... ., Here, a = 27 and d = 24 – 27 = – 3, and, an = 0, , \, an = a + (n – 1)d, , ⇒, 0 = 27 + (n – 1)(– 3), , ⇒, – 3n + 3 = – 27, , ⇒, – 3n = – 27 – 3 = – 30, , ⇒, n = 10, 2. In an Arithmetic Progression, if d = – 4, n = 7,, A [CBSE SQP, 2020-21], an = 4, then find a. , , =, , Sol. We know that, an = a + (n – 1)d, , 4 = a + 6 × (– 4), ½, , a = 28, ½, , [CBSE Marking Scheme, 2020-21], Detailed Solution:, We have, d = – 4, n = 7, and an = 4, , \, an = a + (n – 1)d, , ⇒, 4 = a + (7 – 1)(– 4), , ⇒, 4 = a + 6 (– 4), = a – 24, , ⇒, a = 4 + 24, , ⇒, a = 28, 3. If the first term of an A.P. is p and the common, difference is q, then find its 10th term., R [CBSE Delhi Set-I, 2020], Sol. We have, first term (a) = p,, Common difference (d) = q, and, n = 10, Then,, an = a + (n – 1)d, , ⇒, a10 = p + (10 – 1)q, , ⇒, a10 = p + 9q, 4. Find the common difference of the A.P., , 1 1 −p, ,, ,, p, p, , 1 −2 p, , ............... ., p, , R [CBSE OD Set-I, 2020], , 1 1 − p 1 − 2p, ,, ..., Sol. Given A.P. = ,, p, p, p, , Here, let, , a1 =, , 1, 1−p, and a2 =, p, p, , =, , , , 1−p 1, −, p, p, , 1− p −1, p, −p, p, , = – 1, 5. Find the nth term of the A.P. a, 3a, 5a, ....... ., A [CBSE SQP, 2020-21], , Sol. Given, A.P. = a, 3a, 5a, ..., Here first term,, a = a and d = 3a – a = 2a, , \, nth term = a + (n – 1)d, , = a + (n – 1)2a, , = a + 2na – 2a, , = 2na – a, = (2n – 1)a, 6. How many two digit numbers are divisible by 3 ?, U [CBSE Delhi Set-I, 2019], Sol. Numbers are 12, 15, 18, ..., 99, ½, , \, 99 = 12 + (n – 1) × 3, , ⇒, n = 30, ½, [CBSE Marking Scheme, 2019], , Detailed Solution:, , , Numbers divisible by 3 are 3, 6, 9, 12, 15, -------, 96, 99, , Lowest two digit number divisible by 3 is 12 and, highest two digit number divisible by 3 is 99., , , Hence, the sequence start with 12 ends with 99 and, common difference is 3., , , So, the A.P. will be 12, 15, 18, ----, 96, 99, , Here, a = 12, d = 3, l = 99, , \ , l = a + (n – 1)d, , \ , 99 = 12 + (n – 1)3, , ⇒ , 99 – 12 = 3(n – 1), , ⇒ , , n–1 =, , 87, 3, , , ⇒ , n – 1 = 29, , ⇒ , n = 30, , Therefore, there are 30, two digit numbers divisible, by 3., , 7. Find the sum of the first 10 multiples of 6., , Topper Answer, 2019, , A [CBSE Term, 2019]

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8. If nth term of an A.P. is (2n + 1), what is the sum of, A [CBSE SQP, 2018], its first three terms ?, Sol. Since, a1 = 3, a2 = 5 and a3 = 7, , , , S3 =, , ½, , 3, (3 + 7) = 15, 2, , ½, , [CBSE Marking Scheme, 2018], , Detailed Solution:, , , an = (2n + 1), , ∴, , a1 = 2 × 1 + 1 = 3, , , Since,, Hence,, , , S3 = 15, , Sol. We have,, , 1, m, 1+ m 1, d=, − =1, m, m, a=, , , \, , an =, , 1, + ( n − 1)1, m, , Hence,, , , an =, , 1, +n−1, m, , =, , 9. Which term of the A.P. 8, 14, 20, 26, ....... will be 72, more than its 41st term., A [CBSE OD Set-II, 2017], , , , A [CBSE Delhi Comptt. Set-I, II, III, 2017], , , , l = a3 = 2 × 3 + 1 = 7, n, Sn = [a + l], 2, 3, S3 = [3 + 7], 2, , , n – 1 = 52, , n = 53, 10. Write the nth term of the A.P., 11 11++ m, m 11++ 22m, m, ....., ,,, ,,, ,, ....., m, m m, m, m, m, , [CBSE Comptt. Set-III, 2017], Sol. Given a = 8 and d = 6., Let nth term be 72 more than its 41th term., , ∴, tn – t41 = 72, 8 + (n – 1)6 – (8 + 40 × 6) = 72, , 8 + (n – 1)6 = 320, , (n – 1)6 = 312, , 1 + ( n - 1)m, m, , , 11. If the nth term of the A.P. – 1, 4, 9, 14, .... is 129., Find the value of n., A [CBSE Delhi Comptt. Set-I, II, III, 2017], , Sol. Given, a = – 1 and d = 4 – (– 1) = 5, , an = – 1 + (n – 1) × 5 = 129 ½, or,, (n – 1)5 = 130, , (n – 1) = 26, , n = 27, ½, , [CBSE Marking Scheme, 2017], , 12. What is the common difference of an A.P. in which a21 – a7 = 84 ?, , A [CBSE OD Set-I, II, III, 2017], , Topper Answer, 2017, , 13. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P. ?, C + A [CBSE OD Set-II, 2016], , Topper Answer, 2016

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14. Find the tenth term of the sequence:, , , , , , , , 2 , 8 , 18 , ... ., , U [CBSE SQP, 2016] [CBSE Foreign Set-I, II, III, 2015], , Sol. Given sequence is an A.P., 2 , 8 , 18 , ... =, Hence,, , , or,, , 2 , 2 2 , 3 2 ..., a=, , 2,d=, , =, , 2 + (10 - 1) 2, 2 +9 2, , 15. Is series, , a10 =, , 200, , 3 , 6 , 9 , 12 , ..... an A.P. ? Give reason. , U [CBSE Term-II, 2015], , Sol. Common difference,, d1 =, , , , , Again,, , , 2. Show that (a – b)2, (a2 + b2) and (a + b)2 are in A.P., , =, , d2 =, , 6- 3, 3, , (, , 2 -1, , A [CBSE Delhi Set-I, 2020], 2, , 2, , 2, , Sol. Given: (a – b) , (a + b ) and (a + b)2, Common difference,, , d1 = (a2 + b2) – (a – b)2, = a2 + b2 – (a2 + b2 – 2ab), = a2 + b2 – a2 – b2 + 2ab, = 2ab, and, d2 = (a + b)2 – (a2 + b2), = a2 + b2 +2ab – a2 – b2, = 2ab, , Since,, d1 = d2, Hence, (a – b)2, (a2 + b2) and (a + b)2 are in A.P., , Hence Proved., , , = 10 2, Hence,, , = 990, = 990, = 990 – 100, = 890, 890, n =, = 89, 10, , , , 2 and n = 10, , an = a + (n – 1)d, a10 =, , 110 + 10n – 10, 10n + 100, 10n, 10n, , ), , 9- 6, , 3. Find the sum of first 20 terms of the following A.P.:, 1, 4, 7, 10, ........, , = 3 - 6, d3 = 12 - 9, , = 2 3 - 3, As common differences are not equal., Hence, the given series is not an A.P., [CBSE Marking Scheme, 2015] 1, , Short Answer Type Questions-I, (2 Marks Each), 1. Find the number of natural numbers between 102, and 998 which are divisible by 2 and 5 both., A [CBSE SQP, 2020], , Sol. 110, 120, 130, ........, 990, , an = 990 ⇒ 110 + (n – 1) × 10 = 990, 1, , \, n = 89, 1, [CBSE SQP Marking Scheme, 2020], Detailed Solution:, The number which ends with 0 is divisible by 2 and 5, both., \ Such numbers between 102 and 998 are:, 110, 120, 130, .........., 990., Last term,, an = 990, , a + (n + 1)d = 990, 110 + (n – 1) × 10 = 990, , A [CBSE Delhi Set-II, 2020], , , , Sol. Given A.P.: 1, 4, 7, 10, ..., Here, a = 1, d = 4 – 1 = 3 and n = 20, , \ The sum of first 20 terms,, n, , S20 = [2a + (n – 1)d], 2, =, , 20, [2 × 1 + (20 – 1)3], 2, , = 10(2 + 57), = 10 × 59, = 590, 4. The sum of the first 7 terms of an A.P. is 63 and that, of its next 7 terms is 161. Find the A.P., A [CBSE Delhi Set-III, 2020], , , Sol. Since,, Given,, , Sn =, , n, [2a + (n – 1)d], 2, , S7 = 63, , 7, So,, S7 = [2a +6d], 2, , = 63, or,, 2a + 6d = 18, Now, sum of 14 terms is:, , ...(i)

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S14 = Sfirst 7 terms + Snext 7 terms, = 63 + 161 = 224, 14, \, [2a+ 13d] = 224, 2, , , , ⇒, 2a+ 13d = 32, ...(ii), On subtracting (i) from (ii), we get, (2a +13d) – (2a + 6d) = 32 – 18, , ⇒, 7d = 14, , ⇒, d =2, Putting the value of d in (i), we get, , a =3, Hence, the A.P. will be: 3, 5, 7, 9, ... ., 5. If Sn, the sum of first n terms of an A.P. is given by, Sn = 3n2 – 4n. Find the nth term., A [CBSE Delhi Set-I, 2019], , Sol., a1 = S1 = 3 – 4 = –1, ½, , a2 = S2 – S1, = [3(2)2 – 4(2)] – (–1) = 5 ½, , \, d = a2 – a 1 = 6, ½, Hence, an = –1 + (n – 1) × 6 = 6n – 7 ½, , Alternate method:, , Sn = 3n2 – 4n, , \, Sn – 1 = 3(n – 1)2 – 4(n – 1), = 3n2 – 10n + 7, 1, Hence, an = Sn – Sn – 1, ½, = (3n2 – 4n) – (3n2 – 10n + 7), = 6n – 7, ½, , [CBSE Marking Scheme, 2019], Detailed Solution:, , Given, Sn = 3n2 – 4n, , Put, n = 1, S1 = 3 × 12 – 4 × 1 = – 1, , So, sum of first term of A.P. is – 1., , But sum of first term will be the first term,, , \ First term, a1 = –1, Put n = 2, S2 = 3 × 22 – 4 × 2 = 4, , \ Sum of first two terms is 4., , \ First term + Second term = 4, , \ , –1 + a2 = 4, , ⇒ , a2 = 5, , Hence, Common difference, d = a2 – a1 = 5 – (–1)  = 6, , \ nth term, an = a1 +(n – 1)d, , i.e., , an = –1 + (n – 1)6, , ⇒ , an = 6n – 7, , Therefore, nth term is 6n – 7., , COMMONLY MADE ERROR,  Some students do not know the basic, concepts of arithmetic progression. Many, students try to solve with wrong method., , ANSWERING TIP,  Learn, , the concept of Arithmetic, progression with different examples., , 6. Which term of the A.P. 3, 15, 27, 39,... will be 120, more than its 21st term ?, A [CBSE Delhi Set-I, 2019], , Sol. , , an = a21 + 120, , = (3 + 20 × 12) + 120, = 363, , \, , 1, , 363 = 3 + (n – 1) × 12, , , ⇒, , n = 31, , 1, , st, , Thus, 31 term is 120 more than a21., [CBSE Marking Scheme, 2019], Detailed Solution:, , Given A.P. is: 3, 15, 27, 39, , Here, first term, a = 3 and common difference,, d = 12, , Now, 21st term of A.P. is, , t21 = a + (21 – 1) d, [Q tn = a + (n – 1)d], , \, t21 = 3 +20 × 12 = 243, Therefore, 21st term is 243, We need to calculate term which is 120 more than, 21st term, , i.e., it should be 243 +120 = 363, Therefore,, tn = 363, , \, tn = a + (n – 1)d, , ⇒, 363 = 3+ (n – 1)12, , ⇒, 360 = 12(n – 1), , ⇒, n – 1 = 30, , ⇒, n = 31, , , So, 31st term is 120 more than 21st term., , 7. Find the sum of first 8 multiples of 3. ,  A [CBSE Delhi/OD 2018] [Delhi Comptt. Set-I, 2017], Sol. Here,, , S = 3 + 6 + 9 + 12 + ... + 24, , = 3(1 + 2 + 3 + ... + 8), , , =3×, , = 108, , , 1, , 8×9, 2, , 1, [CBSE Marking Scheme, 2018]

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Detailed Solution:, , Topper Answer, 2019, , 8. Find the 20th term from the last term of the A.P.:, 3, 8, 13, ...... 253., A [CBSE SQP, 2018], Sol. 20th term from the end = l – (n – 1)d, ½, = 253 – 19 × 5, 1, = 158, ½, [CBSE Marking Scheme, 2018], Detailed Solution:, Given A.P.: 3, 8, 13, .......... 253, Here, first term (a) = 3, common difference (d) =, 8 – 3 = 5 and last term (l) = 253, Then, 20th term from the end of the A.P., , = l – (n – 1)d, , = 253 + (20 – 1)5, , = 253 – 95, = 158, 9. Find how many integers between 200 and 500 are, divisible by 8., A [CBSE Delhi Comptt. Set-I, II, III, 2017], , Sol. Integers divisible by 8 are 208, 216, 224, ......, 496., Which is an A.P., Here, a = 208, d = 8 and l = 496, Let the number of terms in A.P. be n., , , , \, , , , an = a + (n – 1)d = l, 208 + (n – 1)d = 496, (n – 1)8 = 496 – 208, 288, n–1 =, 8, , = 36, , n = 36 + 1 = 37, , Hence, no. of required integers divisible by 8 = 37, 10. The fifth term of an A.P. is 26 and its 10th term is 51., Find the A.P. A [CBSE OD Comptt. Set-II, 2017], Sol. Here,, a5 = a + 4d = 26, ...(i) ½, , and, a10 = a + 9d = 51, ...(ii) ½, , Solving Eqns. (i) and (ii), we get, , or,, 5d = 25, , d =5, ½, , and, a =6, Hence, the A.P. is 6, 11, 16 ........, ½, , [CBSE Marking Scheme, 2017], , 11. How many terms of the A.P. − 6 ,, , − 11, 9, , − 5 , − ....., 2, 2, , are needed to give their sum zero., A [CBSE Delhi Comptt. Set-III, 2017], , [CBSE Delhi Set-III, 2016], 11, 1, Sol. Given a = – 6 and d = − − ( − 6 ) =, 2, 2, n, Since,, Sn = [ 2 a + ( n − 1)d ], 2, , Let sum of n terms be zero., \, Sn = 0, n, 1, , or,  2 × − 6 + ( n − 1)  = 0, 2, 2, or,, , n, n 1, −12 + −  = 0, 2 , 2 2, , or,, , n  n 25 , −, =0, 2  2 2 , , or,, n2 – 25n = 0, , n(n – 25) = 0, , n = 25, as n ≠ 0, Hence, required terms are 25., 12. In an A.P. of 50 terms, the sum of the first 10 terms, is 210 and the sum of its last 15 terms is 2565. Find, A [CBSE Foreign Set-III, 2017], the A.P., Sol. Given,, Since,, , , or,, , or,, Since,, and, , Hence,, , S10 = 210, n, Sn = [ 2 a + ( n − 1)d ], 2, 10, (2 a + 9d ) = 210, 2, 2a + 9d = 42, a36 = a + 35d, a50 = a + 49d, , Sum of last 15 terms =, , or,, , ½, ...(i), , 15, (a + 35d + a + 49d ), 2, , 15, (2 a + 84 d ) = 2565, 2, , ½, , , or,, a + 42d = 171, ...(ii) ½, On solving (i) and (ii), we get, , a = 3 and d = 4, ½, , Hence, given A.P. is 3, 7, 11 ......., , [CBSE Marking Scheme, 2017]

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13. How many two digit numbers are divisible by 7 ? , , A [CBSE SQP, 2016], , Sol. Two digit numbers which are divisible by 7 are:, 14, 21, 28, ......, 98., ½, It forms an A.P., Here,, a = 14, d = 7 and an = 98 ½, Since,, an = a + (n – 1)d, , 98 = 14 + (n – 1)7, ½, , 98 – 14 = 7n – 7, , 84 + 7 = 7n, or, , 7n = 91, or, , n = 13, ½, , [CBSE Marking Scheme, 2016], 14. In a certain A.P. 32th term is twice the 12th term., Prove that 70th term is twice the 31st term., A [CBSE Term-II, 2015], Sol. Let the 1st term be a and common difference be d., According to the question, a32 = 2a12, ∴, a + 31d = 2(a + 11d), , , a + 31d = 2a + 22d, , a = 9d, 1, Again,, a70 = a + 69d, = 9d + 69d = 78d, , , a31 = a + 30d, = 9d + 30d = 39d, Hence,, a70 = 2a31, Hence Proved. 1, , [CBSE Marking Scheme, 2015], 15. The 8th term of an A.P. is zero. Prove that its 38th, term is triple of its 18th term., A [CBSE Term-II, 2015], Sol. Given, a8= 0 or, a + 7d = 0 or, a = – 7d, ½, or,, a38 = a + 37d, or,, a38 = – 7d + 37d = 30d, ½, And,, a18 = a + 17d, = – 7d + 17d = 10d, ½, or,, a38 = 30d = 3 × 10d = 3 × a18, , ∴, a38 = 3a18. Hence Proved. ½, , [CBSE Marking Scheme, 2015], 16. The fifth term of an A.P. is 20 and the sum of its, seventh and eleventh terms is 64. Find the common, A [CBSE Foreign Set II, 2015], difference., , [CBSE Term-II, 2015], Sol. Let the first term be a and common difference be d., Then,, a +4d = 20, ...(i) ½, , and a + 6d + a + 10d = 64, , a + 8d = 32, ...(ii) 1, , Solving equations (i) and (ii), we get, , d =3, Hence, common difference, d = 3, ½, , [CBSE Marking Scheme, 2015], 17. Find the middle term of the A.P. 213, 205, 197, ....., A [CBSE Delhi Term, 2015], 37., Sol. Here, a = 213, d = 205 – 213 = – 8 and l = 37, Let the number of terms be n., , , ∴, , or,, or,, , or,, , l = a + (n – 1)d, 37 = 213 + (n – 1)(– 8), 37 – 213 = – 8(n – 1), - 176, n–1 =, = 22, -8, , n = 22 + 1 = 23, 23 + 1, The middle term will be =, = 12th, 2, , , ½, ½, ½, , ∴, a12 = a + (n – 1)d, = 213 + (12 – 1)(– 8), = 213 – 88, = 125, Thus, the middle term will be 125., ½, , [CBSE Marking Scheme, 2015], 18. In an A.P., if S5+ S7 = 167 and S10 = 235, then find, the A.P., where Sn denotes the sum of first n terms., A [CBSE Term-II, 2015], Sol., , n, [ 2 a + (n − 1)d ], 2, , Sn =, , , Given,, , S5 + S7 = 167, 5, 7, Hence, ( 2 a + 4 d ) + ( 2 a + 6d ) = 167, 2, 2, , or, , 24a + 62d = 334, or , 12a + 31d = 167, ....(i) ½, Given,, S10 = 235, or, , 5(2a + 9d) = 235, or , 2a + 9d = 47, ...(ii) ½, , Solving (i) and (ii), wet get, , a = 1 and d = 5, ½, , Hence, A.P. = 1, 6, 11, ...., ½, , [CBSE Marking Scheme, 2015], , Short Answer Type Questions-II, (3 Marks Each), 1, 1, 3, 1. Which term of the A.P. 20, 19 , 18 , 17 , ......, 4, 2, 4, is the first negative term., A [CBSE OD Set-III, 2020], First term, a = 20, 77, 3, and Common difference, d =, − 20 = −, 4, 4, , , , Sol. Here,, , Let, , , , \, , ⇒, , ⇒, , ⇒, , tn < 0, tn = a + (n – 1) d,  3, 20 + (n – 1)  −  < 0,  4, 80 – 3n + 3 < 0, 83 – 3n < 0, 83, n>, 3, , , ⇒, n > 27.6, , ⇒, n = 28, Hence, 28th term will be the first negative term.

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2. Find the middle term of the A.P. 7, 13, 19, ...., 247., U [CBSE OD Set-III, 2020], , Sol. In this A.P., a = 7, d = 13 – 7 = 6, and, , tn = 247, , , , , tn = a + (n – 1)d, , , \, , 247 = 7 + (n – 1)6, , , ⇒, , 6(n – 1) = 240, , , ⇒, , n – 1 = 40, , , ⇒, , n = 41, , Hence,, , , , , =, , 41 + 1, 2, 42, 2, , = 21., Hence, 21st term will be the middle term., , \, , t21 = a + 20d, , = 7 + 20 × 6, = 7 + 120, = 127, , , , , ⇒, , n, [2 × 1 + (n – 1)3] = 287, 2, , , ⇒, , n, [2 + 3n – 3] = 287, 2, , , ⇒, 3n2 – n – 574 = 0, 2, , ⇒ 3n – 42n + 41n – 574 = 0, , ⇒ 3n(n – 14) + 41(n – 14) = 0, , ⇒, (n – 14)(3n + 41) = 0, 41, Either n = 14 or n = –, , it is not possible., 3, , n+1, the middle term =, 2, , =, , 4. Solve the equation: 1 + 4 + 7 + 10 + ... + x = 287., A [CBSE Delhi & OD Set-I, 2020], , Sol. Given, a = 1 and d = 4 – 1 = 3, Let number of terms is the series be n, then, n, , Sn =, [2a + (n – 1)d], 2, , 3. Show that the sum of all terms of an A.P. whose, first term is a, the second term is b and the last, ( a + c )( b + c - 2 a), term is c is equal to, ., 2( b - a), , Thus 14th term, , a14 = a + (14 – 1)d, , ⇒ = 1 + 13 × 3, = 40., 5. If in an A.P., the sum of first m terms is n and the, sum of its first n terms is m, then prove that the, sum of its first (m + n) terms is –(m + n)., A [CBSE OD Set-II, 2020], , Sol. Let 1st term of series be a and common difference be, d, then, , Sm = n[given], m, , ⇒, [2a + (m – 1)d] = n, 2, , A [CBSE OD Set-I, 2020], Sol. Given,, first term, A = a, and, second term = b, , ⇒ common difference, d = b – a, , Last term, l = c, , ⇒, A + (n – 1)d = c, , [By using, l = a + (n – 1)d], , ⇒, a + (n – 1)d = c, , a + (n – 1)(b – a) = c, , ⇒, (b – a)(n – 1) = c – a, c−a, , ⇒, n–1 =, b−a, c−a, +1, , ⇒, n=, b−a, , , ⇒, and, , c−a+b−a, b−a, b + c − 2a, , ⇒, n=, b−a, n, Now, sum =, [A + l], 2, (b + c − 2a), =, [a + c], 2 (b − a), , =, , =, , =, , , ( a + c )( b + c − 2 a ), 2 (b − a), , Hence Proved., , , ⇒, , m[2a + (m – 1)d] = 2n...(i), Sn = m[given], n, [2a + (n – 1)d] = m, 2, , , ⇒, n[2a + (n – 1)d] = 2m...(ii), On subtracting eq. (ii) from eq. (i), , 2(n – m) = 2a(m – n) + d[m2 – n2 – (m – n)], , ⇒, 2(n – m) = 2a(m – n) + d[(m – n)(m + n), , – (m – n)], , ⇒, 2(n – m) = (m – n)[2a + d(m + n – 1)], , ⇒, – 2 = 2a + d(m + n – 1), ...(iii), m+n, Now,, Sm + n =, [2a + (m + n – 1)d], 2, m+n, (– 2), 2, , [from (iii)], , = – (m + n), Hence Proved., 6. Find the sum of all 11 terms of an A.P. whose, A [CBSE OD Set-II, 2020], middle term is 30., Sol. In an A.P. with 11 terms,, 11 + 1, , middle term =, term, 2, , = 6th term, Now, sixth term i.e., a6 = a + (6 – 1)d, i.e.,, a + 5d = 30, ...(i), , [Q middle term i.e., a6 = 30 (given)]

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Now, the sum of 11 terms,, 11, , S11 =, [2a + (11 – 1)d], 2, 11, =, [2a + 10d], 2, 11, =, × 2[a + 5d], 2, = 11 × 30, [from (i)], = 330, 7. If the sum of first m terms of an A.P. is the same as, the sum of its first n terms, show that the sum of its, A [CBSE SQP, 2020], first (m + n) terms is zero., Sol. , , Sm = Sn, m, n, [2a + (m – 1)d] =, [2a + (n – 1d], 1, , ⇒, 2, 2, , ⇒ 2a(m – n) + d(m2 – m – n2+n) = 0, 1, , ⇒, (m – n)[2a + (m + n – 1)d] = 0, 1, or, Sm+n = 0, , [CBSE SQP Marking Scheme, 2020], Detailed Solution:, , Sum of first m terms = Sum of first n terms, , ⇒ , Sm = Sn, m, n, [2a + (m – 1)d] = [2a + (n – 1)d], 2, 2, , , , , m[2a + (m – 1)d] = n[2a + (n – 1)d], , , m[2a + (m – 1)d] – n[2a + (n – 1)d] = 0, , , 2a(m – n) + [m(m – 1) – n (n – 1)]d = 0, , , 2a(m – n) + [m2 – m – n2 + n]d = 0, , 2a(m – n) + [(m – n)(m + n) – (m – n)]d = 0 , , , (m – n)[2a + (m + n – 1)d] = 0, , , Here, (m – n) is not equal to zero., , So, , [2a + (m + n – 1)d] = 0, , Hence,, Sm + n = 0, , 8. If the sum of first four terms of an A.P. is 40 and, that of first 14 terms is 280. Find the sum of its first, A [CBSE Delhi Set-I, 2019], n terms., , Sol. S4 = 40 ⇒ 2(2a + 3d) = 40 ⇒ 2a + 3d = 20, ½, , S14 = 280 ⇒ 7(2a + 13d) = 280 ⇒ 2a + 13d = 40 ½, Solving to get d = 2 , ½, and a = 7 , ½, n, , \, Sn = [14 + (n – 1)2], ½, 2, = n(n + 6) or (n2 + 6n) ½, , [CBSE Marking Scheme, 2019], Detailed Solution:, , Since,, , Sum of n terms of an A.P.,, n, , Sn= [2a + (n – 1)d], 2, [a be the first term and d be, the common difference], , , According to question, S4 = 40, 4, , ⇒ , [2a + (4 – 1)d] = 40, 2, , ⇒ , , ⇒ , , and, , 2[2a+3d] = 40, 2a + 3d = 20, S14 = 280, , , ⇒ , , 14, [2a +(14 – 1)d] = 280, 2, , ...(i), , , ⇒ , 7 (2a + 13d) = 280, , ⇒ , 2a + 13d = 40, ...(ii), , Solving eq. (i) and (ii), we get, a = 7 and d = 2, n, , \ , Sn = [2 × 7 +(n – 1)2], 2, n, = [14+ 2n – 2], , 2, n, = (12 +2n), , 2, , , = 6n+ n2, , Hence, Sum of n terms = 6n + n2, 9. For what value of n, are the nth terms of two A.P.s, 63, 65, 67,.... and 3, 10, 17,.... equal ?, , C + A [CBSE OD Set-III, 2017], Sol. Let a, d and A, D be the 1st term and common, different of the 2 A.P.s respectively., Here,, a = 63, d = 2, , A = 3, D = 7, Given,, an = A n, , ⇒ a + (n – 1) d = A + (n – 1)D, , ⇒ 63 + (n – 1) 2 = 3 + (n – 1) 7, , ⇒, 63 + 2n – 2 = 3 + 7n – 7, , ⇒, 61 + 2n = 7n – 4, , ⇒, 5n = 65, , ⇒, n = 13, , \ When n is 13, the nth terms are equal, , i.e.,, , a13 = A13, th, , 10. If the 10 term of an A.P. is 52 and the 17th term is, 20 more than the 13th term, find A.P., A [CBSE, OD Set-I 2017], , Sol., , a10 = 52, or,, Also, , a + 9d = 52, , ...(i) 1, , a17 – a13 = 20, , a + 16d – (a + 12d) = 20, , , 4d = 20, , , , d =5, , ½, , Substituting, the value of d in (i), we get, , Hence,, , , a =7, A.P. = 7, 12, 17, 22 ....., , 1, ½, , [CBSE Marking Scheme, 2017]

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11. How many terms of an A.P. 9, 17, 25, .... must be taken to give a sum of 636 ?, , A [CBSE OD Set-III, 2017], , Topper Answer, 2017, , 12. Find the sum of n terms of the series, ,  1  2  3,  4 − n  +  4 − n  +  4 − n  + ......, A [CBSE Delhi Set-I, II, III, 2017], , Sol. Let sum of n term be Sn., 1 , 2 , 3, , \ Sn =  4 −  +  4 −  +  4 −  + ......, n , n , n, , up to n terms 1, or, (4 + 4 + 4 + ..... up to n terms), 1 2 3, –  + + + ........ up to n terms, n n n, , or, (4 + 4 + 4 + ..... up to n terms), 1, – (1 + 2 + 3 + ..... up to n terms), n, or, (4 + 4 + 4 + ..... up to n terms), 1, – (1 + 2 + 3 + ..... up to n terms), n, 1 n( n + 1), or, 4 n − ×, 1½, n, 2, , n+1, 7n − 1, or,, =, 4n −, 2, 2, 7n − 1, , Hence, sum of n terms =, ½, 2, , , [CBSE Marking Scheme, 2017], , 13. If the sum of the first 14 terms of an A.P. is 1050 and, its first term is 10, find its 20th term., A [CBSE OD Comptt. Set-III, 2017], , Sol. Given, a = 10, and S14 = 1050, Let the common difference of the A.P. be d., ½, n, Since,, Sn = [ 2 a + ( n − 1)d ], 2, 14, \, S14 =, [ 2 × 10 + (14 − 1) d ], 2, , = 1050, ½, 1050, 20 + 13d =, = 150, 7, , 13d = 130, 130, = 10, , d=, 1, 13, , an = a + (n – 1)d, , a20 = 10 + 19 × 10 = 200, 1, Hence,, a20 = 200, , [CBSE Marking Scheme, 2017], 14. Find the sum of all odd numbers between 0 and 50., A [Delhi Comptt. Set-III, 2017], , Sol. Given, 1 + 3 + 5 + 7 + ..... + 49, Let total odd number of terms be n., an = 1 + (n – 1) × 2 = 49, , (n – 1) × 2 = 49 – 1 = 48, , 1

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n – 1 = 24, n = 24 + 1 = 25, 25, S25 =, (1 + 49 ), 2, , 1, , = 25 × 25, = 625, Hence, sum of odd numbers between 0 and 50, = 625 , 1, , [CBSE Marking Scheme, 2017], 15. If mth term of A.P. is, , 1, 1, , find, and nth term is, n, m, the, A [CBSE Set-I, II, 2017], , sum of first mn terms., , Sol. Let first term of given A.P. be a and common, difference be d., 1, ...(i) ½, , \ am = a + (m – 1)d =, n, 1, and an = a + (n – 1)d =, ...(ii) ½, m, On subtracting (ii) from (i) we get, 1 1 m−n, , (m – n)d = − =, 1, n m, mn, 1, or,, d=, mn, 1, [from (i)], and, a=, mn, Now, , , , , mn  1, 1 , + ( mn − 1),  2·, , Smn =, 2  mn, mn , mn  2, mn, 1 , = 2  mn + mn − mn , mn  1, , +1, Smn =, 2  mn , =, , , , 1, [mn + 1] ., 2, , 1, , [CBSE Marking Scheme, 2017], , 16. Find the sum of all two digit natural numbers, which are divisible by 4., A [Delhi Comptt. Set-II, 2017], Sol. First two digit multiple of 4 is 12 and last is 96, So, a = 12, d = 4 and l = 96, Let nth term be last term = 96, , \, an = a + (n – 1)d = l, , 12 + (n – 1)4 = 96, , n – 1 = 21, , n = 21 + 1 = 22, 22, , Now,, S22 =, [12 + 96 ], 2, , Sol. The series can be written as, (5 + 9 + 13 + .... + 81), + [(– 41) + (– 39) + (– 37) + (– 35) ... (– 5) + (– 3)], For the series (5 + 9 + 13 + ..... + 81), ½, , a =5, , d =4, and, an = 81, Then,, an = 5 + (n – 1)4, = 81, or,, (n – 1)4 = 76, ½, , n = 20, 20, ( 5 + 81), , Sn =, 2, = 860, For series (– 41) + (– 39) + (– 37) + .... + (– 5) + (– 3), , an = – 3, ½, , a = – 41, , d =2, Then,, an = –41 + (n – 1)(2), , \, n = 20, 20, , Sn =, −41 + ( − 3), 2, = – 440, ½, Hence, the sum of the series = 860 – 440, = 420, 1, , [CBSE Marking Scheme, 2017], 18. The ninth term of an A.P. is equal to seven times, the second term and twelfth term exceeds five, times the third term by 2. Find the first term and, A [CBSE SQP, 2016], the common difference., , 1, [mn + 1], 2, , Hence, the sum of first mn terms =, , 17. Find the sum of the following series:, 5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + ...., + (– 5) + 81 + (– 3) A [CBSE Foreign Set-I, 2017], , 1, , 1, , = 11 × 108, = 1188, 1, , [CBSE Marking Scheme, 2017], , Sol. Let the first term of A.P. be a and common difference, be d., Given,, a9 = 7a2, or,, a + 8d = 7(a + d), ...(i) ½, and, a12 = 5a3 + 2, Again,, a + 11d = 5(a + 2d) + 2, ...(ii) 1, From (i),, a + 8d = 7a + 7d, , – 6a + d = 0, ...(iii), From (ii),, a + 11d = 5a + 10d + 2, , – 4a + d = 2, ...(iv), Subtracting (iv) from (iii), we get, , – 2a = – 2, or,, a =1, 1, From (iii),, , –6+d =0, , d =6, ½, Hence, first term = 1 and common difference = 6, [CBSE Marking Scheme, 2016]

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19. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing, A [CBSE Delhi Set-II, 2016], the digits is 594 less than the original number. Find the number., , Topper Answer, 2016, , 20. Divide 56 in four parts in A.P. such that the ratio, of the product of their extremes (1st and 4th) to the, product of middle (2nd and 3rd) is 5 : 6., U [CBSE Foreign Set-I, 2016], Sol. Let the four parts be, a – 3d, a – d, a + d and a + 3d., ∴ a –3d + a – d + a + d + a + 3d = 56, or,, , 4a = 56, , , , a = 14, , Hence, four parts are 14 – 3d, 14 – d, 14 + d and, 14 + 3d., Now, according to question,, , or,, or,, or,, , (14 - 3d )(14 + 3d ), 5, =, (14 - d )(14 + d ), 6, 196 - 9d 2, , 6(196 – 9d2) = 5(196 – d2), 2, , 6 × 196 – 54d = 5 × 196 – 5d, , or, 6 × 196 – 5 × 196 = 54d – 5d, (6 – 5) × 196 = 49d, , 196, =4, 49, , or,, d =±2, , ∴ The four parts are, {14 – 3(± 2)}, {14 –(± 2)}, Hence, first possible division will be 8, 12, 16 and 20., and second possible division will be 20, 16, 12 and 8., 21. The pth, qth and rth terms of an A.P. are a, b and c, respectively. Show that a(q – r) + b(r – p) + c(p – q), = 0., , U [CBSE Foreign Set-II, 2016], , Sol. Let the first term be a' and the common difference, be d., c = a’ + (r – 1)d , , 2, , or,, , d2 =, , , a = a’ + (p – 1)d, b = a’ + (q – 1)d and, , 5, =, 6, , 196 - d 2, , or,, , 2, , 2, , 2, , , , a(q – r) = [a’ + (p – 1)d](q – r), , , , b(r – p) = [a’ + (q – 1)d](r – p), , and, , c(p – q) = [a’ + (r – 1)d](p – q), , 1½, , ½, , , \ a(q – r) + b(r – p) + c(p – q) = a’[q – r + r – p + p –, q] + d[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)], ½

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= a’ × 0 + d[pq – pr + qr – pq + pr – qr + (– q + r – r, + p – p + q)] = 0 , Hence Proved. ½, [CBSE Marking Scheme, 2016], 22. The sum of first n terms of three arithmetic, progressions are S1, S2 and S3 respectively., The first term of each A.P. is 1 and common, differences are 1, 2 and 3 respectively. Prove that, A [CBSE OD Set-III, 2016], S1 + S3 = 2S2., Sol. Since,, , and, or,, Also,, , S1 = 1 + 2 + 3 + .... + n., S2 = 1 + 3 + 5 + ...upto n terms, S3 = 1 + 4 + 7 + ...upto n terms, n( n + 1), S1 =, ½, 2, S2 =, , n, [2 × 1 + (n – 1)2], 2, , =, , n, [2n] = n2, 2, , and, , n, [2 × 1 + (n – 1)3], 2, , S3 =, , ½, , =, , n( 3n - 1), 2, , ½, , Now,, , n( n + 1) n( 3n - 1), +, 2, 2, , ½, , S1 + S3 =, , =, , n[n + 1 + 3n - 1], 2, , =, , n[ 4 n], 2, , , a1 = S1 = 5, ½, , a2 = S2 – S1 = 13 – 5 = 8, ½, , d = a2 – a 1 = 8 – 5 = 3, ½, Now, A.P. is 5, 8, 11, ...... ., , nth term, an = a + (n – 1)d, = 5 + (n – 1)3, = 3n + 2, Hence,, a20 = 3 × 20 + 2, , a20 = 62, ½, , [CBSE Marking Scheme, 2015], 24. Prove that the nth term of an A.P. can not be n2 + 1., Justify your answer., [CBSE Term-II, 2015], Sol. Let nth term of A.P.,, , an = n2 + 1, Putting the values of n = 1, 2, 3, ......, we get, , a1 = 1 2 + 1 = 2, , a2 = 2 2 + 1 = 5, , a3 = 32 + 1 = 10, 1, The obtained sequence, = 2, 5, 10, 17, ........., Their common difference, , = a2 – a 1 ¹ a 3 – a 2 ¹ a 4 – a 3, , or,, 5 – 2 ¹ 10 – 5 ¹ 17 – 10, , \, 3 ¹5¹7, 1, Since the common difference are not equal., Hence, n2 + 1 is not a form of nth term of an A.P. 1, , [CBSE Marking Scheme, 2015], 25. If Sn denotes, the sum of the first n terms of an A.P., prove that S12 = 3(S8 – S4)., A [CBSE Delhi, Set-I, 2015], , 2, , = 2n, = 2S2, Hence Proved. 1, [CBSE Marking Scheme, 2016], 23. If the sum of the first n terms of an A.P. is, 1, [3n2 + 7n], then find its nth term. Hence write its, 2, 20th term., , , A [CBSE Term-II, Set-II, 2015], , [CBSE SQP, 2016], , Sol. , , Sn =, , 1, [3n2 + 7n], 2, , , , S1 =, , 1, [3 × (1)2 + 7(1)], 2, , , , =, , 1, [3 + 7], 2, , =, , 1, × 10 = 5, 2, , , , 1, [3 (2)2 + 7 × 2], 2, , S2 =, , , , =, , =, , ½, , 1, [12 + 14], 2, , , S12 = 6[2a + 11d], = 12a + 66d, ...(i) 1, , S8 = 4[2a + 7d], = 8a + 28d, ½, , and, S4 = 2[2a + 3d], = 4a + 6d, ½, Then,, 3(S8 – S4) = 3[(8a + 28d) – (4a + 6d)], = 3[4a + 22d], = 12a + 66d, , From equation (i) and (ii),, , S12 = 3(S8 – S9), 1, , [CBSE Marking Scheme, 2015], , , , 1, × 26, 2, , = 13, , Sol. Let a be the first term and d be the common, difference., n, Since,, Sn = [ 2 a + ( n − 1)d ], 2, , , ½, , 26. The 14th term of an A.P. is twice its 8th term. If, the 6th term is – 8, then find the sum of its first 20, A [CBSE OD Set-I, 2015], terms., [CBSE Foreign Set-I, II, 2015], , Sol. Let first term be a and common difference be d., , Here,, a14 = 2a8, , or,, a + 13d = 2(a + 7d)

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a + 13d = 2a + 14d, , a =–d, ...(i) ½, , Again,, a6 = – 8, or,, a + 5d = – 8, ...(ii) ½, , Solving (i) and (ii), we get, , a = 2, d = – 2, ½, 20, [ 2 × 2 + ( 20 - 1)( -2 )] ½, S20 =, 2, , = 10[4 + 19 × (– 2)], = 10(4 – 38), = 10 × (– 34) = – 340, 1, , [CBSE Marking Scheme, 2015], , Long Answer Type Questions, (5 Marks Each), 1. The sum of four consecutive numbers in A.P. is, 32 and the ratio of the product of the first and, , , , last term to the product of two middle terms is, 7 : 15. Find the numbers. U [CBSE Delhi Set-I, 2020], [CBSE Delhi & OD 2018], Sol. Let the four consecutive terms of A.P. be, (a – 3d), (a – d), (a + d) and (a + 3d)., 1, By given conditions, , a – 3d + a – d + d + a + 3d = 32, , ⇒, 4a = 32, , ⇒, a =8, 1, ( a - 3d )( a + 3d ), 7, =, 1, And, ( a - d )( a + d ), 15, , a 2 - 9d 2, 7, =, 2, 2, 15, a, d, , , d2 = 4, , d = ±2, 1, Hence, the numbers are 2, 6, 10 and 14 or 14, 10, 6, and 2. , 1, , [CBSE Marking Scheme, 2018], , 2. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the, (m + n)th term of the A.P. is zero., [CBSE Term-I, II, III, 2019], , Topper Answer, 2019

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3. The first term of an A.P. is 3, the last term is 83 and, the sum of all its terms is 903. Find the number of, terms and the common difference of the A.P., , , [CBSE Delhi Set-II, 2019], , Sol. Here a = 3, an = 83 and Sn = 903, Therefore, 83 = 3 + (n – 1)d, , ⇒, (n – 1)d = 80, n, Also, 903 = [2a + (n – 1)d], 2, =, , n, (6 + 80), 2, , 1, ...(i) 1, , (using (i)), , = 43n, 1+½, , ⇒, n = 21, and from eq. (i), d =4, 1+½, [CBSE Marking Scheme, 2019], , Detailed Solution:, , Given:, , First term, a = 3, , Last term, an = 83, , Sum of n terms, Sn= 903, n, , Since,, Sn = ( a + an ), 2, n, , ⇒ , 903 = (3 + 83), 2, , ⇒ , ⇒ , , ⇒ , , Now,, , 903 = 43n, 903, n=, 43, n = 21, n, Sn = [2a + (n – 1)d], 2, , , ⇒ , , 903 =, , 21, [2 × 3 + (21 – 1)d], 2, , , ⇒ , 903 = 21(3 + 10d), , ⇒ , 3 + 10d = 43, , ⇒ , 10d = 40, , ⇒ , d=4, , Hence, the common difference is 4., , COMMONLY MADE ERROR,  Some students fail to find the value of n, , as they get confused between the nth term, and last term., , ANSWERING TIP,  Understand the formulae related to given, condition and use them to solve the, problems., , 4. An A.P. consists of 50 terms of which 3rd term is 12, and last term is 106. Find the 29th term. , U [CBSE SQP, 2018], Sol. Given, n = 50, a3 = 12 and a50 = 106, Then, a + 2d = 12, 1, and, a + 49d = 106, 1, On solving, we get d = 2 and a = 8, 1+1, Now,, a29 = a + 28d, = 8 + 28 × 2, = 64, 1, , [CBSE Marking Scheme, 2018], , 5. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th, A [CBSE OD Set-III, 2017] [CBSE OD Set-I, 2016], terms., , Topper Answer, 2017

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6. The ratio of the sums of first m and first n terms of, an A.P. is m2 : n2. Show that the ratio of its mth and, nth terms is (2m – 1) : (2n – 1)., [CBSE Delhi Set-I, 2017], Sol. Let first term of given A.P. be a and common, difference be d also sum of first m and first n terms, be Sm and Sn respectively., Sm, m2, , \, = 2, 1, Sn, n, m, [ 2 a + (m − 1)d ] m2, , or, 2, = 2, n, n , 2 a + ( n − 1)d ], [, 2, ⇒, , , 1, , 2, 2 a + ( m − 1)d, = m ×n =m, 2 a + ( n − 1)d, n2 m n, , , ⇒ m(2a + (n – 1)d) = n[2a + (m – 1)d], , ⇒ 2am + nmd – md = 2an + nmd – nd, , ⇒, (n – m)d = 2a(n – m), , ⇒, d = 2a, am, a + ( m − 1)d, =, , Now, an, a + ( n − 1)d, =, , a + ( m − 1) × 2 a, a + ( n − 1) × 2 a, , , a + 2ma − 2 a, 2ma − a, or,, =, a, +, 2, na, −, 2, a, 2na − a, , =, , 1, , , or,, , , Now,, , Hence,, [Q d = 2a], , , , a( 2m − 1), a( 2n − 1), , 1, 1, 7. If the pth term of an A.P. is, and qth term is, ., q, p, Prove that the sum of first pq term of the A.P. is, [CBSE Delhi Set-III, 2017], , Sol. Try yourself similar to Q.No. 15 of SATQ-II., , 2(a + 17d) = 3(a + 10d), a = 4d, 5, (2 a + 4d ), S5, 2, =, S10, 10, [2 a + 9d ], 2, , ...(i), , Putting the value of a = 4d, we get, 5, (8 d + 4 d ), S5, 2, or,, =, 5 (8 d + 9 d ), S10, , , , , , = (2m – 1) : (2n – 1), 2, , Hence Proved., ,  pq+1 ,  2  ., , , , 8. If the ratio of the 11th term of an A.P. to its 18th term, is 2 : 3, find the ratio of the sum of the first five, term to the sum of its first 10 terms., [Delhi Comptt. Set-I, II, III, 2017], a11, a + 10d 2, Sol. Since,, =, =, a + 17 d 3, a18, , 6, 12d, =, 17, 34 d, S5 : S10 = 6 : 17, , 9. An A.P. consists of 37 terms. The sum of the three, middle most terms is 225 and the sum of the last, three terms is 429. Find the A.P. [CBSE SQP, 2017], Sol. Let the middle most terms of the A.P. be (a – d),, a and (a + d)., Given , a – d + a + a + d = 225, or,, , 3a = 225, , or,, , a = 75, , , and the middle term =, , 37 + 1, = 19th term, 2, , , \ A.P. is, (a – 18d), ....., (a – 2d), (a – d), a, (a + d), (a + 2d), .... ,, (a + 18d), , Sum of last three terms, (a + 18d) + (a + 17d) + (a + 16d) = 429

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or,, or,, , 3a + 51d = 429, 225 + 51d = 429 or, d = 4, , , , First term, a1 = a – 18d = 75 – 18 × 4 = 3., , , a2 = 3 + 4 = 7, , Hence, A.P. = 3, 7, 11, ........ , 147., , 1, , 10. The sum of three numbers in A.P. is 12 and sum of, their cubes is 288. Find the numbers., A [Delhi Set-III, 2016], Sol. Let the three numbers in A.P. be a – d, a and a + d., Then, their sum i.e., 3a = 12, 1, or,, a =4, Also, (4 – d)3 + 43 + (4 + d)3 = 288, 1, or, 64 – 48d + 12d2 – d3 + 64 + 64 + 48d + 12d2 + d3, = 288, or,, 24d2 + 192 = 288, 1, or,, d2 = 4, , ∴, d =±2, 1, Hence, the numbers are 2, 4 and 6, or 6, 4 and 2. 1, , [CBSE Marking Scheme, 2016], , 11. Find the value of a, b and c such that the numbers, a, 7, b, 23 and c are in A.P., U [CBSE Term-II, 2015], , Sol. Since, a, 7, b, 23 and c are in A.P., Let the common difference be d, , ∴, a+d =7, ...(i) ½, and, a + 3d = 23, ...(ii) ½, From (i) and (ii), we get, , a = – 1 and d = 8, 1, Again,, b = a + 2d, , b =–1+2×8, or,, b = – 1 + 16, or,, b = 15, 1, , ∴, c = a + 4d, = – 1 + 4 × 8, = – 1 + 32, , c = 31, 1, , ∴a = – 1, b = 15 and c = 31, 1, , [CBSE Marking Scheme, 2015]