CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set 17

Question. Find the sum of the following APs:
(i) 2, 7, 12, ....., to 10 terms.
(ii) –37, –33, –29, ....., to 12 terms.
(iii) 0.6, 1.7, 2.8, ....., to 100 terms.
(iv) \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots \), to 11 terms.
Answer: (i) Here, \( a = 2, d = 7 – 2 = 5, n = 10 \)
\( S_n = \frac{n}{2} [2a + (n – 1) d] \)
\( \implies S_{10} = \frac{10}{2} [2 \times 2 + (10 – 1) 5] \)
\( \implies S_{10} = 5 [4 + 45] = 5 \times 49 = 245 \)
(ii) Here, \( a = –37, d = –33 – (–37) = 4, n = 12 \)
\( S_n = \frac{n}{2} [2a + (n – 1) d] \)
\( \implies S_{12} = \frac{12}{2} [2 \times (–37) + (12 – 1) 4] \)
\( \implies S_{12} = 6[–74 + 44] = 6 \times (– 30) = –180 \)
(iii) Here, \( a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100 \)
\( S_n = \frac{n}{2} [2a + (n – 1) d] \)
\( \implies S_{100} = \frac{100}{2} [2 \times 0.6 + (100 – 1) 1.1] \)
\( \implies S_{100} = 50 [1.2 + 108.9] = 50 \times 110.1 = 5505 \)
(iv) Here, \( a = \frac{1}{15}, d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}, n = 11 \)
\( S_n = \frac{n}{2} [2a + (n – 1) d] \)
\( \implies S_{11} = \frac{11}{2} [2 \times \frac{1}{15} + (11 - 1) \frac{1}{60}] \)
\( \implies S_{11} = \frac{11}{2} [\frac{2}{15} + \frac{10}{60}] = \frac{11}{2} [\frac{2}{15} + \frac{1}{6}] = \frac{11}{2} [\frac{4 + 5}{30}] \)
\( \implies S_{11} = \frac{11}{2} \times \frac{9}{30} = \frac{11 \times 3}{20} = \frac{33}{20} \)

Question. Find the sums given below:
(i) \( 7 + 10\frac{1}{2} + 14 + \dots + 84 \)
(ii) \( 34 + 32 + 30 + \dots + 10 \)
(iii) \( –5 + (–8) + (– 11) + \dots + (– 230) \)
Answer: (i) Here, \( a = 7, d = 10.5 - 7 = 3.5 = \frac{7}{2}, a_n = 84 \)
\( a_n = a + (n – 1) d \)
\( \implies 84 = 7 + (n – 1) \frac{7}{2} \)
\( \implies 77 = (n – 1) \frac{7}{2} \)
\( \implies n – 1 = 22 \)
\( \implies n = 23 \)
\( S_n = \frac{n}{2} [a + l] \)
\( \implies S_{23} = \frac{23}{2} [7 + 84] = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046\frac{1}{2} \)
(ii) Here, \( a = 34, d = 32 – 34 = –2, a_n = 10 \)
\( a_n = a + (n – 1) d \)
\( \implies 10 = 34 + (n – 1) (–2) \)
\( \implies –24 = –2 (n – 1) \)
\( \implies n – 1 = 12 \)
\( \implies n = 13 \)
\( S_{13} = \frac{13}{2} [34 + 10] = \frac{13 \times 44}{2} = 13 \times 22 = 286 \)
(iii) Here, \( a = – 5, a_n = – 230, d = – 8 – (–5) = – 3 \)
\( a_n = a + (n – 1) d \)
\( \implies – 230 = –5 + (n – 1) (–3) \)
\( \implies – 225 = –3 (n – 1) \)
\( \implies n – 1 = 75 \)
\( \implies n = 76 \)
\( S_{76} = \frac{76}{2} [– 5 + (– 230)] = 38 \times (– 235) = –8930 \)

Question. In an AP:
(i) given \( a = 5, d = 3, a_n = 50 \), find \( n \) and \( S_n \).
(ii) given \( a = 7, a_{13} = 35 \), find \( d \) and \( S_{13} \).
(iii) given \( a_{12} = 37, d = 3 \), find \( a \) and \( S_{12} \).
(iv) given \( a_3 = 15, S_{10} = 125 \), find \( d \) and \( a_{10} \).
(v) given \( d = 5, S_9 = 75 \), find \( a \) and \( a_9 \).
(vi) given \( a = 2, d = 8, S_n = 90 \), find \( n \) and \( a_n \).
(vii) given \( a = 8, a_n = 62, S_n = 210 \), find \( n \) and \( d \).
(viii) given \( a_n = 4, d = 2, S_n = –14 \), find \( n \) and \( a \).
(ix) given \( a = 3, n = 8, S = 192 \), find \( d \).
(x) given \( l = 28, S = 144 \), and there are total 9 terms. Find \( a \).
Answer: (i) \( a_n = a + (n - 1)d \)
\( \implies 50 = 5 + (n - 1)3 \)
\( \implies 45 = 3(n - 1) \)
\( \implies 15 = n - 1 \)
\( \implies n = 16 \)
\( S_{16} = \frac{16}{2}[5 + 50] = 8 \times 55 = 440 \)
(ii) \( a_{13} = a + 12d \implies 35 = 7 + 12d \)
\( \implies 28 = 12d \)
\( \implies d = \frac{7}{3} \)
\( S_{13} = \frac{13}{2}[7 + 35] = \frac{13 \times 42}{2} = 273 \)
(iii) \( a_{12} = a + 11d \implies 37 = a + 33 \)
\( \implies a = 4 \)
\( S_{12} = \frac{12}{2}[4 + 37] = 6 \times 41 = 246 \)
(iv) \( a + 2d = 15 \implies a = 15 - 2d \)
\( S_{10} = \frac{10}{2}[2a + 9d] = 125 \)
\( \implies 5[2(15 - 2d) + 9d] = 125 \)
\( \implies 30 - 4d + 9d = 25 \implies 5d = -5 \implies d = -1 \)
\( a = 15 - 2(-1) = 17 \); \( a_{10} = a + 9d = 17 + 9(-1) = 8 \)
(v) \( S_9 = \frac{9}{2}[2a + 8d] = 75 \)
\( \implies \frac{9}{2}[2a + 40] = 75 \implies 9a + 180 = 75 \)
\( \implies 9a = -105 \implies a = -\frac{35}{3} \)
\( a_9 = a + 8d = -\frac{35}{3} + 40 = \frac{85}{3} \)
(vi) \( S_n = \frac{n}{2}[4 + (n - 1)8] = 90 \)
\( \implies n[2 + 4n - 4] = 90 \implies n(4n - 2) = 90 \implies 2n^2 - n - 45 = 0 \)
\( \implies (n - 5)(2n + 9) = 0 \implies n = 5 \)
\( a_5 = 2 + 4(8) = 34 \)
(vii) \( S_n = \frac{n}{2}[8 + 62] = 210 \implies 35n = 210 \implies n = 6 \)
\( a_6 = 8 + 5d = 62 \implies 5d = 54 \implies d = 10.8 \)
(viii) \( a_n = a + (n-1)2 = 4 \implies a = 6 - 2n \)
\( S_n = \frac{n}{2}[a + 4] = -14 \implies n(6 - 2n + 4) = -28 \)
\( \implies n(10 - 2n) = -28 \implies n^2 - 5n - 14 = 0 \)
\( \implies (n - 7)(n + 2) = 0 \implies n = 7 \)
\( a = 6 - 2(7) = -8 \)
(ix) \( S_8 = \frac{8}{2}[2(3) + 7d] = 192 \implies 4[6 + 7d] = 192 \)
\( \implies 6 + 7d = 48 \implies 7d = 42 \implies d = 6 \)
(x) \( S_9 = \frac{9}{2}[a + 28] = 144 \implies a + 28 = 32 \implies a = 4 \)

Question. How many terms of AP: 9, 17, 25, ..... must be taken to give a sum of 636?
Answer: Here, \( a = 9, d = 17 - 9 = 8, S_n = 636 \)
\( S_n = \frac{n}{2} [2a + (n – 1) d] \)
\( \implies 636 = \frac{n}{2} [2 \times 9 + (n – 1) 8] \)
\( \implies 1272 = n [18 + 8n – 8] = n [10 + 8n] \)
\( \implies 8n^2 + 10n - 1272 = 0 \implies 4n^2 + 5n – 636 = 0 \)
\( \implies 4n^2 + 53n – 48n – 636 = 0 \)
\( \implies n(4n + 53) - 12(4n + 53) = 0 \)
\( \implies (n - 12)(4n + 53) = 0 \)
\( \implies n = 12 \) (since \( n \) must be a positive integer)
Hence, 12 terms must be taken.

Question. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer: Given, \( a = 5, a_n = 45, S_n = 400 \)
\( S_n = \frac{n}{2} [a + a_n] \)
\( \implies 400 = \frac{n}{2} [5 + 45] \implies 400 = 25n \)
\( \implies n = 16 \)
Now, \( a_n = a + (n - 1) d = 45 \)
\( \implies 5 + (16 - 1) d = 45 \implies 15d = 40 \)
\( \implies d = \frac{40}{15} = \frac{8}{3} \)

Question. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer: Here, \( a = 17, a_n = 350, d = 9 \)
\( a_n = a + (n – 1) d \)
\( \implies 350 = 17 + (n – 1) 9 \)
\( \implies 333 = (n – 1) 9 \)
\( \implies n – 1 = 37 \implies n = 38 \)
\( S_{38} = \frac{38}{2} [17 + 350] = 19 \times 367 = 6973 \)

Question. Find the sum of first 22 terms of an AP in which \( d = 7 \) and 22nd term is 149.
Answer: Given, \( d = 7, a_{22} = 149 \)
\( a_{22} = a + 21d = 149 \implies a + 21 \times 7 = 149 \)
\( \implies a + 147 = 149 \implies a = 2 \)
\( S_{22} = \frac{22}{2} [a + a_{22}] = 11 [2 + 149] = 11 \times 151 = 1661 \)

Question. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer: Given, \( a_2 = 14 \) and \( a_3 = 18 \)
\( d = a_3 - a_2 = 18 - 14 = 4 \)
\( a = a_2 - d = 14 - 4 = 10 \)
\( S_{51} = \frac{51}{2} [2(10) + (51 - 1)4] = \frac{51}{2} [20 + 200] = \frac{51 \times 220}{2} = 51 \times 110 = 5610 \)

Question. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first \( n \) terms.
Answer: \( S_7 = 49 \implies \frac{7}{2}[2a + 6d] = 49 \implies a + 3d = 7 \) -- (i)
\( S_{17} = 289 \implies \frac{17}{2}[2a + 16d] = 289 \implies a + 8d = 17 \) -- (ii)
Subtracting (i) from (ii): \( 5d = 10 \implies d = 2 \).
From (i), \( a + 3(2) = 7 \implies a = 1 \).
\( S_n = \frac{n}{2} [2(1) + (n – 1) 2] = \frac{n}{2} [2n] = n^2 \)

Question. Show that \( a_1, a_2, ....., a_n, ..... \) form an AP where \( a_n \) is defined as below:
(i) \( a_n = 3 + 4n \)
(ii) \( a_n = 9 – 5n \)
Also, find the sum of the first 15 terms in each case.
Answer: (i) \( a_n = 3 + 4n \)
\( a_1 = 7, a_2 = 11, a_3 = 15 \dots \)
Common difference \( d = 11 - 7 = 4 \). Hence it's an AP.
\( S_{15} = \frac{15}{2} [2 \times 7 + 14 \times 4] = \frac{15}{2} [14 + 56] = 525 \)
(ii) \( a_n = 9 – 5n \)
\( a_1 = 4, a_2 = -1, a_3 = -6 \dots \)
Common difference \( d = -1 - 4 = -5 \). Hence it's an AP.
\( S_{15} = \frac{15}{2} [2 \times 4 + 14 \times (-5)] = \frac{15}{2} [8 - 70] = -465 \)

Question. If the sum of the first \( n \) terms of an AP is \( 4n – n^2 \), what is the first term (that is \( S_1 \))? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the \( n \)th terms. [DoE]
Answer: \( S_n = 4n – n^2 \)
\( a_1 = S_1 = 4(1) - 1^2 = 3 \)
Sum of first two terms \( S_2 = 4(2) - 2^2 = 4 \)
Second term \( a_2 = S_2 - S_1 = 4 - 3 = 1 \)
Common difference \( d = a_2 - a_1 = 1 - 3 = -2 \)
\( a_3 = a + 2d = 3 + 2(-2) = -1 \)
\( a_{10} = a + 9d = 3 + 9(-2) = -15 \)
\( a_n = a + (n - 1)d = 3 + (n - 1)(-2) = 5 - 2n \)

Question. Find the sum of the first 40 positive integers divisible by 6.
Answer: Numbers are 6, 12, 18, ....., 240
\( a = 6, d = 6, n = 40 \)
\( S_{40} = \frac{40}{2} [6 + 240] = 20 \times 246 = 4920 \)

Question. Find the sum of the first 15 multiples of 8.
Answer: Numbers are 8, 16, 32, ....., 120
\( a = 8, d = 8, n = 15 \)
\( S_{15} = \frac{15}{2} [8 + 120] = \frac{15 \times 128}{2} = 960 \)

Question. Find the sum of the odd numbers between 0 and 50. [DoE]
Answer: Odd numbers are 1, 3, 5, ....., 49
\( a = 1, d = 2, a_n = 49 \implies 1 + (n - 1)2 = 49 \implies n = 25 \)
\( S_{25} = \frac{25}{2} [1 + 49] = 25 \times 25 = 625 \)

Question. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc. the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Answer: \( a = \text{Rs. } 200, d = \text{Rs. } 50, n = 30 \)
\( S_{30} = \frac{30}{2} [2 \times 200 + 29 \times 50] = 15 [400 + 1450] = 15 \times 1850 = \text{Rs. } 27,750 \)

Question. A sum of Rs. 700 is to be used to give seven cash prizes to students for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer: \( S_7 = 700, d = -20, n = 7 \)
\( S_7 = \frac{7}{2} [2a + 6(-20)] = 700 \implies a - 60 = 100 \implies a = 160 \)
The prizes are Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60 and Rs. 40.

Question. In a school, students thought of planting 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 the same as the class, in which they are studying, e.g. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? [AI 2014]
Answer: Trees per section form AP: 1, 2, 3, ....., 12
\( S_{12} = \frac{12}{2} [1 + 12] = 6 \times 13 = 78 \)
Total trees for 3 sections \( = 78 \times 3 = 234 \)

SELECT EXEMPLAR PROBLEMS

Question. In an AP, if \( a = 3.5, d = 0, n = 101 \), then \( a_n \) will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Answer: (b) 3.5

Question. The list of numbers –10, –6, –2, 2, ... is
(a) an AP with \( d = –16 \)
(b) an AP with \( d = 4 \)
(c) an AP with \( d = –4 \)
(d) not an AP
Answer: (b) an AP with \( d = 4 \)

Question. Two APs have the same common difference. The first term of one of these is –1 and that of the other is –8. Then the difference between their 4th terms is
(a) –1
(b) –8
(c) 7
(d) –9
Answer: (c) 7

Question. Find the sum of two middle most terms of the A.P. \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4\frac{1}{3} \)
Answer: \( a = -\frac{4}{3}, d = \frac{1}{3}, a_n = \frac{13}{3} \implies n = 18 \).
Middle terms are 9th and 10th.
\( a_9 + a_{10} = 2a + 17d = 2(-\frac{4}{3}) + 17(\frac{1}{3}) = \frac{9}{3} = 3 \).

Question. Find the sum of the integers between 100 and 200 that are (i) divisible by 9. (ii) not divisible by 9.
Answer: (i) Numbers: 108, ..., 198. \( a = 108, d = 9, a_n = 198 \implies n = 11 \).
\( S_{11} = \frac{11}{2}[108 + 198] = 1683 \).
(ii) Total sum (101-199): \( a = 101, d = 1, a_n = 199 \implies n = 99 \).
\( S_{99} = \frac{99}{2}[101 + 199] = 14850 \).
Sum not divisible by 9 \( = 14850 - 1683 = 13167 \).

Question. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
Answer: \( n = 27 \), middle flag is 14th.
Distance for 13 flags on each side: \( S = 2 \times 2 [2 + 4 + \dots + 26] = 4 \times \frac{13}{2} [2 + 26] = 728 \text{ m} \).
Max distance carrying flag \( = 13 \times 2 = 26 \text{ m} \).

Question. Justify whether it is true to say that –1, \( -\frac{3}{2}, -2, -\frac{5}{2} \dots \) forms an AP as \( a_2 – a_1 = a_3 – a_2 \).
Answer: Yes. \( -\frac{3}{2} - (-1) = -0.5 \) and \( -2 - (-\frac{3}{2}) = -0.5 \). Common difference is constant.

Question. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?
Answer: \( a_n - A_n = (a + (n-1)d) - (A + (n-1)d) = a - A = 2 - 7 = -5 \). The difference depends only on the first terms.

Question. The taxi fare after each km, when the fare is Rs. 15 for the first km and Rs. 8 for each additional km, does not form an AP as the total fare (in Rs.) after each km is 15, 8, 8, 8, ..... Is the statement true? Give reasons.
Answer: False. Total fare is 15, 23, 31, ... which is an AP with \( d = 8 \).

Question. Find the sum of the AP whose 7th term is 24 less than the 11th term, first term being 12.
Answer: \( a = 12, a_{11} - a_7 = 24 \implies 4d = 24 \implies d = 6 \).
\( a_{20} = 12 + 19(6) = 126 \).
(Note: Solution for 20th term was given in OCR, sum would require \( n \)).

Question. Determine \( k \) so that \( k^2 + 4k + 8, 2k^2 + 3k + 6 \) and \( 3k^2 + 4k + 4 \) are three consecutive terms of an AP.
Answer: \( 2(2k^2 + 3k + 6) = (k^2 + 4k + 8) + (3k^2 + 4k + 4) \)
\( \implies 4k^2 + 6k + 12 = 4k^2 + 8k + 12 \implies 2k = 0 \implies k = 0 \).

Question. If \( S_n \) denotes the sum of first \( n \) terms of an AP, prove that \( S_{12} = 3(S_8 – S_4) \)
Answer: \( S_8 = 4(2a + 7d), S_4 = 2(2a + 3d) \).
\( 3(S_8 - S_4) = 3(8a + 28d - 4a - 6d) = 3(4a + 22d) = 12a + 66d = S_{12} \).

Question. How many terms of the AP: –15, –13, –11, ... are needed to make the sum –55? Explain the reason for double answer.
Answer: \( a = -15, d = 2, S_n = -55 \).
\( \frac{n}{2}[-30 + (n-1)2] = -55 \implies n^2 - 16n + 55 = 0 \implies (n-5)(n-11) = 0 \).
\( n = 5 \) or \( n = 11 \).
Double answer because sum of terms from 6th to 11th is zero.