Home › Class 10 › NCERT Solutions › Chapter 5: Arithmetic Progressions › Exercise 5.3 Show all solutions Try each question first, then open its solution. Our own step-by-step solutions, set out for step marks.
Exercise 5.3 questions and solutions
Exercise 5.3, Question 1
Find the sum.
(i) \(2, 7, 12, \ldots\) to 10 terms
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\(a = 2,\ d = 5\). \(S_{10} = 5[4 + 9(5)] = 5 \times 49 = 245\). Answer: \(245\)
(ii) \(-37, -33, -29, \ldots\) to 12 terms
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\(a = -37,\ d = 4\). \[\begin{aligned}S_{12} &= 6[-74 + 11(4)] \\ &= 6 \times (-30) \\ &= -180\end{aligned}\] Answer: \(-180\)
(iii) \(0.6, 1.7, 2.8, \ldots\) to 100 terms
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\(a = 0.6,\ d = 1.1\). \[\begin{aligned}S_{100} &= 50[1.2 + 99(1.1)] \\ &= 50 \times 110.1 \\ &= 5505\end{aligned}\] Answer: \(5505\)
(iv) \(\tfrac{1}{15}, \tfrac{1}{12}, \tfrac{1}{10}, \ldots\) to 11 terms
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\[\begin{gathered}a = \tfrac1{15}, \\ d = \tfrac1{12} - \tfrac1{15} = \tfrac1{60}\end{gathered}\] \[\begin{aligned}S_{11} &= \tfrac{11}{2}\left[\tfrac{2}{15} + \tfrac{10}{60}\right] \\ &= \tfrac{11}{2} \times \tfrac{3}{10}\end{aligned}\] \(= \tfrac{33}{20}\). Answer: \(\tfrac{33}{20}\)
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Exercise 5.3, Question 2
Find the sum.
(i) \(7 + 10\tfrac12 + 14 + \cdots + 84\)
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\(a = 7,\ d = 3.5,\ l = 84\). Number of terms: \(84 = 7 + 3.5(n-1) \Rightarrow n = 23\). \[\begin{aligned}S &= \tfrac{23}{2}(7 + 84) \\ &= \tfrac{2093}{2}\end{aligned}\] Answer: \(1046\tfrac12\)
(ii) \(34 + 32 + 30 + \cdots + 10\)
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\(a = 34,\ d = -2,\ l = 10\): \(10 = 34 - 2(n-1) \Rightarrow n = 13\). \(S = \tfrac{13}{2}(34 + 10) = 286\). Answer: \(286\)
(iii) \(-5 + (-8) + (-11) + \cdots + (-230)\)
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\(a = -5,\ d = -3,\ l = -230\): \(-230 = -5 - 3(n-1) \Rightarrow n = 76\). \(S = 38(-5 - 230) = -8930\). Answer: \(-8930\)
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Exercise 5.3, Question 3
In an AP, \(a\) is the first term, \(d\) the common difference, \(a_n\) the nth term, \(l\) the last term and \(S_n\) the sum of \(n\) terms.
(i) \(a = 5,\ d = 3,\ a_n = 50\). Find \(n\) and \(S_n\).
Show solution
\(5 + 3(n-1) = 50 \Rightarrow n = 16\). \(S_{16} = 8(5 + 50) = 440\). Answer: \(n = 16,\ S_n = 440\)
(ii) \(a = 7,\ a_{13} = 35\). Find \(d\) and \(S_{13}\).
Show solution
\(7 + 12d = 35 \Rightarrow d = \tfrac73\). \(S_{13} = \tfrac{13}{2}(7 + 35) = 273\). Answer: \(d = \tfrac73,\ S_{13} = 273\)
(iii) \(a_{12} = 37,\ d = 3\). Find \(a\) and \(S_{12}\).
Show solution
\(a + 11(3) = 37 \Rightarrow a = 4\). \(S_{12} = 6(4 + 37) = 246\). Answer: \(a = 4,\ S_{12} = 246\)
(iv) \(a_3 = 15,\ S_{10} = 125\). Find \(d\) and \(a_{10}\).
Show solution
\(a + 2d = 15\) and \[\begin{aligned}&S_{10} = 5(2a + 9d) = 125 \\ \Rightarrow\ &2a + 9d = 25\end{aligned}\] From the first, \(2a = 30 - 4d\); so \(30 + 5d = 25 \Rightarrow d = -1,\ a = 17\). \(a_{10} = 17 - 9 = 8\). Answer: \(d = -1,\ a_{10} = 8\)
(v) \(d = 5,\ S_9 = 75\). Find \(a\) and \(a_9\).
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\[\begin{aligned}&S_9 = \tfrac92(2a + 40) = 75 \\ \Rightarrow\ &2a + 40 = \tfrac{50}{3}\end{aligned}\] \(a = -\tfrac{35}{3}\), and \(a_9 = a + 40 = \tfrac{85}{3}\). Answer: \(a = -\tfrac{35}{3},\ a_9 = \tfrac{85}{3}\)
(vi) \(a = 2,\ d = 8,\ S_n = 90\). Find \(n\) and \(a_n\).
Show solution
\[\begin{aligned}&\tfrac n2[4 + 8(n-1)] = 90 \\ \Rightarrow\ &4n^2 - 2n - 90 = 0 \\ \Rightarrow\ &2n^2 - n - 45 = 0\end{aligned}\] \((n - 5)(2n + 9) = 0\); \(n\) is a positive integer, so \(n = 5\). \(a_5 = 2 + 4(8) = 34\). Answer: \(n = 5,\ a_n = 34\)
Where marks slip: Reject the negative or fractional value of \(n\) and say why.
(vii) \(a = 8,\ a_n = 62,\ S_n = 210\). Find \(n\) and \(d\).
Show solution
\[\begin{aligned}&S_n = \tfrac n2(8 + 62) = 35n = 210 \\ \Rightarrow\ &n = 6\end{aligned}\] \[\begin{aligned}&62 = 8 + 5d \\ \Rightarrow\ &d = \tfrac{54}{5}\end{aligned}\] Answer: \(n = 6,\ d = \tfrac{54}{5}\)
(viii) \(a_n = 4,\ d = 2,\ S_n = -14\). Find \(n\) and \(a\).
Show solution
\(a + 2(n-1) = 4 \Rightarrow a = 6 - 2n\). \[\begin{aligned}S_n &= \tfrac n2(a + 4) \\ &= \tfrac n2(10 - 2n) \\ &= n(5 - n) \\ &= -14\end{aligned}\] \[\begin{aligned}&n^2 - 5n - 14 = 0 \\ \Rightarrow\ &(n - 7)(n + 2) = 0 \\ \Rightarrow\ &n = 7\end{aligned}\], so \(a = -8\). Answer: \(n = 7,\ a = -8\)
Where marks slip: \(n = -2\) is rejected: the number of terms is positive.
(ix) \(a = 3,\ n = 8,\ S = 192\). Find \(d\).
Show solution
\[\begin{aligned}&192 = 4(6 + 7d) \\ \Rightarrow\ &48 = 6 + 7d \\ \Rightarrow\ &d = 6\end{aligned}\] Answer: \(d = 6\)
(x) \(l = 28,\ S = 144\), and there are 9 terms. Find \(a\).
Show solution
\[\begin{aligned}&144 = \tfrac92(a + 28) \\ \Rightarrow\ &a + 28 = 32 \\ \Rightarrow\ &a = 4\end{aligned}\] Answer: \(a = 4\)
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Exercise 5.3, Question 4
How many terms of \(9, 17, 25, \ldots\) give a sum of \(636\)?
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\(a = 9,\ d = 8\): \(\tfrac n2[18 + 8(n-1)] = n(4n + 5) = 636\). \[\begin{aligned}&4n^2 + 5n - 636 = 0 \\ \Rightarrow\ &(n - 12)(4n + 53) = 0\end{aligned}\] \(n = 12\) (the other root is negative). Answer: \(12\) terms
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Exercise 5.3, Question 5
First term \(5\), last term \(45\), sum \(400\). Find the number of terms and \(d\).
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\[\begin{aligned}&400 = \tfrac n2(5 + 45) = 25n \\ \Rightarrow\ &n = 16\end{aligned}\] \(45 = 5 + 15d \Rightarrow d = \tfrac83\). Answer: \(16\) terms, \(d = \tfrac83\)
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Exercise 5.3, Question 6
First and last terms \(17\) and \(350\), \(d = 9\). How many terms, and what is their sum?
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\[\begin{aligned}&350 = 17 + 9(n-1) \\ \Rightarrow\ &n - 1 = 37 \\ \Rightarrow\ &n = 38\end{aligned}\] \(S = 19(17 + 350) = 6973\). Answer: \(38\) terms, sum \(6973\)
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Exercise 5.3, Question 7
Find the sum of the first 22 terms of an AP with \(d = 7\) and 22nd term \(149\).
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\(a + 21(7) = 149 \Rightarrow a = 2\). \(S_{22} = 11(2 + 149) = 1661\). Answer: \(1661\)
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Exercise 5.3, Question 8
Find the sum of the first 51 terms of an AP whose 2nd and 3rd terms are \(14\) and \(18\).
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\(d = 18 - 14 = 4\), so \(a = 10\). \[\begin{aligned}S_{51} &= \tfrac{51}{2}[20 + 50(4)] \\ &= 51 \times 110 \\ &= 5610\end{aligned}\] Answer: \(5610\)
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Exercise 5.3, Question 9
The sums of the first 7 and first 17 terms of an AP are \(49\) and \(289\). Find the sum of the first \(n\) terms.
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\[\begin{aligned}&S_7 = \tfrac72(2a + 6d) = 49 \\ \Rightarrow\ &a + 3d = 7\end{aligned}\] \[\begin{aligned}&S_{17} = \tfrac{17}{2}(2a + 16d) = 289 \\ \Rightarrow\ &a + 8d = 17\end{aligned}\] So \(d = 2,\ a = 1\), and \(S_n = \tfrac n2[2 + 2(n-1)] = n^2\). Answer: \(S_n = n^2\)
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Exercise 5.3, Question 10
Show that \(a_1, a_2, \ldots, a_n\) form an AP, and find the sum of the first 15 terms.
(i) \(a_n = 3 + 4n\)
Show solution
\[\begin{aligned}a_{n+1} - a_n &= [3 + 4(n+1)] - [3 + 4n] \\ &= 4\end{aligned}\], the same for every \(n\): an AP with \(d = 4\). \(a_1 = 7\), \(a_{15} = 63\); \(S_{15} = \tfrac{15}{2}(7 + 63) = 525\). Answer: AP with \(d = 4\); \(S_{15} = 525\)
(ii) \(a_n = 9 - 5n\)
Show solution
\(a_{n+1} - a_n = -5\) for every \(n\): an AP with \(d = -5\). \(a_1 = 4\), \(a_{15} = -66\); \(S_{15} = \tfrac{15}{2}(4 - 66) = -465\). Answer: AP with \(d = -5\); \(S_{15} = -465\)
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Exercise 5.3, Question 11
The sum of the first \(n\) terms of an AP is \(4n - n^2\). Find the first term, the sum of the first two terms, the second term, and the 3rd, 10th and nth terms.
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\(S_1 = 3\), so the first term is \(3\). \(S_2 = 8 - 4 = 4\). \(a_2 = S_2 - S_1 = 1\); \(a_3 = S_3 - S_2 = 3 - 4 = -1\). \[\begin{aligned}a_{10} &= S_{10} - S_9 \\ &= -60 - (-45) \\ &= -15\end{aligned}\] \[\begin{aligned}a_n &= S_n - S_{n-1} \\ &= (4n - n^2) - [4(n-1) - (n-1)^2] \\ &= 5 - 2n\end{aligned}\] Answer: \[\begin{gathered}a_1 = 3, \\ S_2 = 4, \\ a_2 = 1, \\ a_3 = -1, \\ a_{10} = -15, \\ a_n = 5 - 2n\end{gathered}\]
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Exercise 5.3, Question 12
Find the sum of the first 40 positive integers divisible by \(6\).
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\(6, 12, \ldots, 240\): \(a = 6,\ l = 240\). \(S_{40} = 20(6 + 240) = 4920\). Answer: \(4920\)
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Exercise 5.3, Question 13
Find the sum of the first 15 multiples of \(8\).
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\(8, 16, \ldots, 120\). \(S_{15} = \tfrac{15}{2}(8 + 120) = 960\). Answer: \(960\)
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Exercise 5.3, Question 14
Find the sum of the odd numbers between \(0\) and \(50\).
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\(1, 3, \ldots, 49\): 25 terms. \(S = \tfrac{25}{2}(1 + 49) = 625\). Answer: \(625\)
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Exercise 5.3, Question 15
A late-completion penalty is ₹200 for day 1, ₹250 for day 2, ₹300 for day 3, and so on. What is the penalty for 30 days' delay?
Show solution
An AP with \(a = 200,\ d = 50\). \[\begin{aligned}S_{30} &= 15[400 + 29(50)] \\ &= 15 \times 1850 \\ &= 27750\end{aligned}\] Answer: ₹27 750
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Exercise 5.3, Question 16
₹700 is shared as seven prizes, each ₹20 less than the one before. Find each prize.
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\[\begin{aligned}&S_7 = \tfrac72[2a + 6(-20)] = 700 \\ \Rightarrow\ &2a - 120 = 200 \\ \Rightarrow\ &a = 160\end{aligned}\] Answer: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40
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Exercise 5.3, Question 17
Each of the 3 sections of classes I to XII plants as many trees as its class number. How many trees in all?
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One section of each class plants \(1 + 2 + \cdots + 12 = 78\). Three sections: \(3 \times 78 = 234\). Answer: \(234\) trees
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Exercise 5.3, Question 18
A spiral is made of 13 successive semicircles of radii \(0.5, 1.0, 1.5, 2.0, \ldots\) cm. Find its length (take \(\pi = \tfrac{22}{7}\)).
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A semicircle of radius \(r\) has length \(\pi r\). Total \[\begin{aligned}= \pi(0.5 + 1 + \cdots + 6.5) &= \pi \times \tfrac{13}{2}(0.5 + 6.5) \\ &= 45.5\pi\end{aligned}\] \(45.5 \times \tfrac{22}{7} = 143\). Answer: \(143\) cm
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Exercise 5.3, Question 19
200 logs are stacked with 20 in the bottom row, 19 in the next, 18 in the next, and so on. How many rows, and how many logs in the top row?
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\[\begin{aligned}&S_n = \tfrac n2[40 - (n-1)] = 200 \\ \Rightarrow\ &n^2 - 41n + 400 = 0\end{aligned}\] \((n - 16)(n - 25) = 0\). With 25 rows the top row would have \(20 - 24 = -4\) logs, so \(n = 16\). Top row: \(20 - 15 = 5\). Answer: \(16\) rows, \(5\) logs in the top row
Where marks slip: Both roots are positive integers; the question's context rules out 25.
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Exercise 5.3, Question 20
In a potato race a bucket is 5 m from the first potato and the 10 potatoes are 3 m apart in a line. A runner fetches each potato to the bucket in turn. Total distance run?
Show solution
Round trips: \[2 \times 5, 2 \times 8, 2 \times 11, \ldots\], an AP with \(a = 10,\ d = 6\), 10 terms. \(S_{10} = 5[20 + 9(6)] = 370\). Answer: \(370\) m
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Done the NCERT exercises? The board paper asks more
Arithmetic Progressions has 39 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
Arithmetic Progressions in our sample papers: Sample paper 1 (questions 5, 20, 32, 36) · Sample paper 2 (questions 5, 6, 32) · Sample paper 3 (questions 6, 7, 32, 38) · Sample paper 4 (questions 6, 7, 27, 32) · Sample paper 5 (questions 6, 32).
Also useful: free MCQs and case studies for Arithmetic Progressions · Class 10 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan
← Exercise 5.2 All Chapter 5 exercises Exercise 5.4 → Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in . Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log .