NCERT Exemplar Solutions Class 10 Maths Arithmetic Progressions — Exercise 5.3 Q19–35

NCERT Exemplar Problems, Class 10 Mathematics — Arithmetic Progressions, Exercise 5.3, questions 19 to 35.

This is the second half of Exercise 5.3 — the first eighteen questions, on terms, are on their own page. Everything here is about sums, and three tools cover all of it:

$$S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}\left(a + l\right), \qquad a_n = S_n – S_{n-1}$$

The last of these is the one students meet least and the exemplar uses most. If a question hands you a formula for $S_n$ — Questions 24 and 25 — then subtracting consecutive sums recovers the terms, and no other route is needed.

Two structural facts save a great deal of work:

• The middle term shortcut. If an AP has an odd number of terms, $S_n = n \times (\text{middle term})$. Question 29 is one line this way.
• A sum equation is quadratic in $n$, so it can have two valid answers. Question 32 asks you to explain the double answer rather than discard one, and the explanation is the most interesting thing on this page.

Key insight. $S_n$ is a quadratic in $n$ with no constant term. That single fact explains why $a_n = S_n – S_{n-1}$ is linear (hence an AP), why sum questions can have two answers, and why a formula like $S_n = n(4n+1)$ determines the AP completely.

Question 19

Find the sum of the two middle most terms of the AP $-\dfrac43$, $-1$, $-\dfrac23$, …, $4\dfrac13$.

Solution. Here $a = -\frac43$ and $d = -1 – \left(-\frac43\right) = \frac13$. The last term is $4\frac13 = \frac{13}{3}$, so

$$-\frac43 + (n-1)\frac13 = \frac{13}{3} \implies (n-1) = 13 + 4 = 17 \implies n = 18$$

With $n$ even there are two middle terms, the $\frac{n}{2}$-th and the $\left(\frac{n}{2}+1\right)$-th — here the 9th and 10th:

$$a_9 = -\frac43 + 8\left(\frac13\right) = \frac43, \qquad a_{10} = -\frac43 + 9\left(\frac13\right) = \frac53$$

$$a_9 + a_{10} = \frac43 + \frac53 = \frac93 = 3$$

Question 20

The first term of an AP is $-5$ and the last term is $45$. If the sum of the terms of the AP is $120$, then find the number of terms and the common difference.

Solution. With both end terms known, use $S_n = \frac{n}{2}(a + l)$ — no $d$ needed for the first part:

$$120 = \frac{n}{2}(-5 + 45) = 20n \implies n = 6$$

Now $d$ from the last term:

$$45 = -5 + (6-1)d \implies 5d = 50 \implies d = 10$$

Question 21

Find the sum.

(i) $1 + (-2) + (-5) + (-8) + \dots + (-236)$

An AP with $a = 1$, $d = -3$, last term $-236$. First find how many terms:

$$1 + (n-1)(-3) = -236 \implies 3(n-1) = 237 \implies n = 80$$

$$S_{80} = \frac{80}{2}\left(1 + (-236)\right) = 40 \times (-235) = -9400$$

(ii) $\left(4 – \dfrac1n\right) + \left(4 – \dfrac2n\right) + \left(4 – \dfrac3n\right) + \dots$ upto $n$ terms

The common difference is $-\frac1n$ — constant, because $n$ is fixed while the numerator advances. Summing directly is cleanest: there are $n$ copies of $4$, minus $\frac1n$ times $1 + 2 + \dots + n$:

$$S = 4n – \frac1n \times \frac{n(n+1)}{2} = 4n – \frac{n+1}{2} = \frac{8n – n – 1}{2} = \frac{7n – 1}{2}$$

(iii) $\dfrac{a-b}{a+b} + \dfrac{3a-2b}{a+b} + \dfrac{5a-3b}{a+b} + \dots$ to $11$ terms

Every term has the same denominator, so factor it out and work with the numerators $a-b$, $3a-2b$, $5a-3b$, … The $k$-th numerator is $(2k-1)a – kb$, so summing $k = 1$ to $11$:

$$\sum_{k=1}^{11}\left[(2k-1)a – kb\right] = a\left(2 \times 66 – 11\right) – b(66) = 121a – 66b = 11\left(11a – 6b\right)$$

using $1+2+\dots+11 = 66$. Hence

$$S_{11} = \frac{11\left(11a – 6b\right)}{a + b}$$

Question 22

Which term of the AP $-2$, $-7$, $-12$, … will be $-77$? Find the sum of this AP upto the term $-77$.

Solution. $a = -2$, $d = -5$.

$$-2 + (n-1)(-5) = -77 \implies 5(n-1) = 75 \implies n = 16$$

So $-77$ is the 16th term, and

$$S_{16} = \frac{16}{2}\left(-2 + (-77)\right) = 8 \times (-79) = -632$$

Question 23

If $a_n = 3 – 4n$, show that $a_1$, $a_2$, $a_3$, … form an AP. Also find $S_{20}$.

Solution. Compute the general difference:

$$a_{n+1} – a_n = \left[3 – 4(n+1)\right] – \left[3 – 4n\right] = -4$$

This is free of $n$, so the sequence is an AP with $d = -4$. Its first term is $a_1 = 3 – 4 = -1$.

$$S_{20} = \frac{20}{2}\left[2(-1) + 19(-4)\right] = 10\left[-2 – 76\right] = 10(-78) = -780$$

Question 24

In an AP, if $S_n = n(4n + 1)$, find the AP.

Solution. Recover the terms with $a_n = S_n – S_{n-1}$. First,

$$S_n = 4n^2 + n$$

$$S_{n-1} = 4(n-1)^2 + (n-1) = 4n^2 – 8n + 4 + n – 1 = 4n^2 – 7n + 3$$

$$a_n = S_n – S_{n-1} = \left(4n^2 + n\right) – \left(4n^2 – 7n + 3\right) = 8n – 3$$

So $a_1 = 5$, $a_2 = 13$, $a_3 = 21$, and the AP is

$$5, \ 13, \ 21, \ \dots$$

with $d = 8$. (Check: $S_1 = 1 \times 5 = 5$ ✓ and $S_2 = 2 \times 9 = 18 = 5 + 13$ ✓.)

Question 25

In an AP, if $S_n = 3n^2 + 5n$ and $a_k = 164$, find the value of $k$.

Solution. Same method:

$$S_{n-1} = 3(n-1)^2 + 5(n-1) = 3n^2 – 6n + 3 + 5n – 5 = 3n^2 – n – 2$$

$$a_n = S_n – S_{n-1} = \left(3n^2 + 5n\right) – \left(3n^2 – n – 2\right) = 6n + 2$$

Setting $a_k = 164$:

$$6k + 2 = 164 \implies 6k = 162 \implies k = 27$$

Question 26

If $S_n$ denotes the sum of the first $n$ terms of an AP, prove that $S_{12} = 3\left(S_8 – S_4\right)$.

Solution. Write all three out with the standard formula.

$$S_{12} = \frac{12}{2}\left[2a + 11d\right] = 6\left(2a + 11d\right) = 12a + 66d$$ $$S_8 = \frac{8}{2}\left[2a + 7d\right] = 4\left(2a + 7d\right) = 8a + 28d$$ $$S_4 = \frac{4}{2}\left[2a + 3d\right] = 2\left(2a + 3d\right) = 4a + 6d$$

Then

$$S_8 – S_4 = (8a + 28d) – (4a + 6d) = 4a + 22d$$ $$3\left(S_8 – S_4\right) = 12a + 66d = S_{12}$$

which is the required identity, and it holds for every $a$ and $d$.

There is a reading of it worth noticing: $S_8 – S_4$ is the sum of terms $5$ to $8$ — the middle third of the first twelve. The identity says the whole twelve sum to three times that middle third, which is true because in an AP the three blocks $1$–$4$, $5$–$8$, $9$–$12$ are themselves equally spaced.

Question 27

Find the sum of first $17$ terms of an AP whose $4$th and $9$th terms are $-15$ and $-30$ respectively.

Solution. The difference of the two given terms gives $d$:

$$a_9 – a_4 = 5d = -30 – (-15) = -15 \implies d = -3$$

Then from $a_4 = -15$: $\ a + 3(-3) = -15$, so $a = -6$.

$$S_{17} = \frac{17}{2}\left[2(-6) + 16(-3)\right] = \frac{17}{2}\left[-12 – 48\right] = \frac{17}{2}(-60) = -510$$

Question 28

If the sum of the first $6$ terms of an AP is $36$ and that of the first $16$ terms is $256$, find the sum of the first $10$ terms.

Solution. Two equations in $a$ and $d$:

$$S_6 = 3\left(2a + 5d\right) = 36 \implies 2a + 5d = 12$$ $$S_{16} = 8\left(2a + 15d\right) = 256 \implies 2a + 15d = 32$$

Subtracting, $10d = 20$, so $d = 2$ and $2a = 12 – 10 = 2$, giving $a = 1$.

$$S_{10} = \frac{10}{2}\left[2(1) + 9(2)\right] = 5(20) = 100$$

Question 29

Find the sum of all the $11$ terms of an AP whose middle most term is $30$.

Solution. With $11$ terms — an odd number — there is a single middle term, the 6th, so $a_6 = a + 5d = 30$. Now

$$S_{11} = \frac{11}{2}\left[2a + 10d\right] = 11\left(a + 5d\right) = 11 \times 30 = 330$$

The general fact, worth carrying: for an AP with an odd number of terms, $S_n = n \times (\text{middle term})$, because the middle term is the average of them all.

Question 30

Find the sum of last ten terms of the AP $8$, $10$, $12$, …, $126$.

Solution. The neat way is to read the AP backwards: the last ten terms are the first ten of the reversed AP, which starts at $126$ and has common difference $-2$.

$$S = \frac{10}{2}\left[2(126) + 9(-2)\right] = 5\left[252 – 18\right] = 5(234) = 1170$$

(Forwards: $a = 8$, $d = 2$ and $126 = 8 + 2(n-1)$ gives $n = 60$; the last ten are terms $51$ to $60$, running from $108$ to $126$, and $\frac{10}{2}(108 + 126) = 1170$ ✓.)

Question 31

Find the sum of first seven numbers which are multiples of $2$ as well as of $9$.

Solution. A number that is a multiple of both $2$ and $9$ is a multiple of their LCM. Since $2$ and $9$ are coprime,

$$\operatorname{LCM}(2, 9) = 18$$

The first seven multiples of $18$ are $18$, $36$, $54$, $72$, $90$, $108$, $126$ — an AP with $a = 18$, $d = 18$.

$$S_7 = \frac{7}{2}\left(18 + 126\right) = \frac{7}{2}(144) = 504$$

Or simply $18(1+2+\dots+7) = 18 \times 28 = 504$.

Question 32

How many terms of the AP $-15$, $-13$, $-11$, … are needed to make the sum $-55$? Explain the reason for the double answer.

Solution. With $a = -15$, $d = 2$:

$$S_n = \frac{n}{2}\left[-30 + 2(n-1)\right] = \frac{n}{2}\left(2n – 32\right) = n(n – 16)$$

Setting this to $-55$:

$$n^2 – 16n + 55 = 0 \implies (n – 5)(n – 11) = 0 \implies n = 5 \text{ or } n = 11$$

Both answers are genuine. The reason is that the AP starts negative and turns positive: its terms are

$$-15, \ -13, \ -11, \ -9, \ -7, \ -5, \ -3, \ -1, \ 1, \ 3, \ 5, \ \dots$$

The first five sum to $-55$. Terms $6$ through $11$ are $-5, -3, -1, 1, 3, 5$, which cancel in pairs and add to zero — so adding them changes nothing, and the first eleven also sum to $-55$.

This is what a quadratic in $n$ looks like when a run of terms happens to cancel, and it is why a sum equation can have two admissible roots where a term equation never does.

Question 33

The sum of the first $n$ terms of an AP whose first term is $8$ and the common difference is $20$ is equal to the sum of the first $2n$ terms of another AP whose first term is $-30$ and the common difference is $8$. Find $n$.

Solution. Write both sums, watching that the second uses $2n$ terms:

$$\frac{n}{2}\left[16 + 20(n-1)\right] = \frac{2n}{2}\left[-60 + 8(2n-1)\right]$$

$$\frac{n}{2}\left(20n – 4\right) = n\left(16n – 68\right)$$

Since $n \ne 0$, divide through by $n$:

$$10n – 2 = 16n – 68 \implies 6n = 66 \implies n = 11$$

Question 34

Kanika was given her pocket money on Jan 1st, 2008. She puts ₹1 on Day 1, ₹2 on Day 2, ₹3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent ₹204 of her pocket money, and found that at the end of the month she still had ₹100 with her. How much was her pocket money for the month?

Solution. January has 31 days, so the amounts put into the piggy bank are $1, 2, 3, \dots, 31$ — an AP with $a = 1$, $d = 1$.

$$S_{31} = \frac{31}{2}(1 + 31) = \frac{31 \times 32}{2} = 496$$

The pocket money was divided three ways: into the piggy bank, spent, and left over.

$$\text{pocket money} = 496 + 204 + 100 = ₹800$$

The “$31$” is the whole question — using $30$ days gives $465$ and an answer of ₹769, which is wrong for no visible reason.

Question 35

Yasmeen saves ₹32 during the first month, ₹36 in the second month and ₹40 in the third month. If she continues to save in this manner, in how many months will she save ₹2000?

Solution. The monthly savings form an AP with $a = 32$, $d = 4$, and the question asks for $n$ with $S_n = 2000$:

$$\frac{n}{2}\left[64 + 4(n-1)\right] = 2000$$ $$n\left(4n + 60\right) = 4000$$ $$4n^2 + 60n – 4000 = 0 \implies n^2 + 15n – 1000 = 0$$

Factorising ($40$ and $-25$ have product $-1000$ and sum $15$):

$$(n + 40)(n – 25) = 0 \implies n = 25 \text{ or } n = -40$$

A number of months cannot be negative, so $n = 25$.

Common mistakes

• Taking one middle term when $n$ is even. Question 19 has $18$ terms, so there are two middle ones, the 9th and 10th. An odd $n$ gives one (Question 29); an even $n$ gives two.
• Forgetting that $S_{n-1}$ needs $n-1$ substituted everywhere. In Questions 24 and 25 the whole expression, both the $n^2$ term and the linear one, has to be shifted. Expanding $(n-1)^2$ carefully is most of the work.
• Discarding the second root in Question 32. Both $5$ and $11$ are correct, and the question explicitly asks why. The answer is that the intervening terms sum to zero — check that before assuming a root is spurious.
• Using $2n$ in only one place in Question 33. The second AP contributes $\frac{2n}{2}\left[2a + (2n-1)d\right]$: the $2n$ appears both as the number of terms and inside the bracket.
• Using 30 days in Question 34. January has 31. The sum $1 + 2 + \dots + 31 = 496$, not $465$.
• Not simplifying before solving in Question 32. $S_n = n(n-16)$ is much easier to set equal to $-55$ than the unsimplified bracket, and the factorisation $(n-5)(n-11)$ then appears without the quadratic formula.
• Summing forwards in Question 30. Reversing the AP turns “last ten terms” into “first ten terms” and removes the need to count all sixty.

Practise next

• Exemplar Exercise 5.4 — the long-answer sums, including the multiples-of-2-or-5 counting and the loan instalments.
• Exemplar Exercise 5.3 Q1–18 for the $n$-th term work these sums assume.
• NCERT textbook Exercise 5.3 — twenty questions on $S_n$, including the savings and instalment problems of Questions 34 and 35.
• NCERT textbook Exercise 5.4 for the harder optional set.