NCERT Exemplar Solutions Class 11 Maths Sequences and Series — Short and Long Answer Type Q1–16

NCERT Exemplar Problems, Class 11 Mathematics — Sequences and Series, Short and Long Answer Type questions 1 to 16. All 16 questions solved.

Most of these questions run on three formulae. For an A.P. with first term $a$ and common difference $d$, $T_n = a + (n-1)d$ and $S_n = \frac n2[2a + (n-1)d]$; for a G.P. with ratio $r$, $T_n = ar^{n-1}$; and for any sequence, $T_n = S_n – S_{n-1}$. The word problems (2, 3, 5, 6, 8, 9) only need you to spot the A.P. hidden in the story. The proofs (1, 4, 10, 13 to 16) are algebra with a trick: telescoping in 10 and 14, and a clean choice of variables in 13 and 16.

Question 7 is ambiguous, and the printed key’s answer corresponds to a different count from the literal one. It is explained where it occurs.

Key insight. Write everything in terms of the first term and the common difference (or ratio), then let the given conditions fix those two unknowns. In the proofs, look for a quantity that telescopes: rationalising in Question 10 and the identity for $\sin(\theta_{k+1}-\theta_k)$ in Question 14 each turn a long sum into two end terms.

Question 1

The first term of an A.P. is $a$, and the sum of the first $p$ terms is zero. Show that the sum of its next $q$ terms is $\dfrac{-a(p+q)q}{p-1}$. [Hint: required sum $= S_{p+q} – S_p$.]

Solution. The “next $q$ terms” are terms $p+1$ to $p+q$, so their sum is $S_{p+q} – S_p$, and $S_p = 0$ makes it simply $S_{p+q}$. First use $S_p = 0$ to find $d$:

$$\frac p2[2a + (p-1)d] = 0 \;\Rightarrow\; 2a + (p-1)d = 0 \;\Rightarrow\; d = \frac{-2a}{p-1}.$$

Now

$$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] = \frac{p+q}{2}\left[2a – \frac{2a(p+q-1)}{p-1}\right] = (p+q)\,a\cdot\frac{(p-1)-(p+q-1)}{p-1} = \frac{-a(p+q)q}{p-1}.$$

Question 2

A man saved Rs 66000 in 20 years. In each succeeding year after the first year he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year?

Solution. The yearly savings form an A.P. with common difference $200$, and the total over 20 years is $S_{20} = 66000$. Let the first-year saving be $a$:

$$\frac{20}{2}[2a + 19 \times 200] = 66000 \;\Rightarrow\; 2a + 3800 = 6600 \;\Rightarrow\; a = 1400.$$

Question 3

A man accepts a position with an initial salary of Rs 5200 per month. He will receive an automatic increase of Rs 320 in the very next month and each month thereafter. (a) Find his salary for the tenth month. (b) What is his total earnings during the first year?

Solution. Monthly salaries form an A.P. with $a = 5200$ and $d = 320$.

Question 3 (a)

$T_{10} = 5200 + 9 \times 320 = 5200 + 2880 = 8080$.

Question 3 (b)

A year has 12 months, so we need $S_{12}$:

$$S_{12} = \frac{12}{2}[2 \times 5200 + 11 \times 320] = 6[10400 + 3520] = 6 \times 13920 = 83520.$$

Question 4

If the $p$th and $q$th terms of a G.P. are $q$ and $p$ respectively, show that its $(p+q)$th term is $\left(\dfrac{q^p}{p^q}\right)^{\frac{1}{p-q}}$.

Solution. Let the first term be $A$ and the ratio $r$. Then

$$Ar^{p-1} = q, \qquad Ar^{q-1} = p.$$

Dividing, $r^{p-q} = \dfrac qp$. The $(p+q)$th term is $Ar^{p+q-1} = (Ar^{q-1})\,r^{p} = p\,r^p$. Now raise $r^{p-q} = q/p$ to the power $\frac{p}{p-q}$ to get $r^p = \left(\dfrac qp\right)^{\frac{p}{p-q}}$. Hence

$$T_{p+q} = p\left(\frac qp\right)^{\frac{p}{p-q}} = \left(p^{\,p-q}\right)^{\frac1{p-q}}\left(\frac{q^p}{p^p}\right)^{\frac1{p-q}} = \left(\frac{p^{p-q}\,q^p}{p^p}\right)^{\frac1{p-q}} = \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.$$

Question 5

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter, he made two more frames than he made the day before. How many days did it take him to finish the job?

Solution. Daily output is an A.P. with $a = 5$, $d = 2$, and the total after $n$ days must be $192$:

$$\frac n2[10 + 2(n-1)] = 192 \;\Rightarrow\; n(n+4) = 192 \;\Rightarrow\; n^2 + 4n – 192 = 0 \;\Rightarrow\; (n+16)(n-12) = 0.$$

A number of days cannot be negative, so $n = 12$. (Check: $5 + 7 + \dots + 27 = \frac{12}{2}(5+27) = 192$.)

Question 6

We know the sum of the interior angles of a triangle is $180^\circ$. Show that the sums of the interior angles of polygons with $3, 4, 5, 6, \dots$ sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.

Solution. A polygon with $n$ sides can be cut into $n-2$ triangles from one vertex, so its angle sum is $(n-2) \times 180^\circ$. For $n = 3, 4, 5, \dots$ this gives $180^\circ, 360^\circ, 540^\circ, \dots$: each term is $180^\circ$ more than the one before, so the sums form an A.P. with $a = 180$ and $d = 180$. For $n = 21$:

$$19 \times 180 = 3420.$$

Question 7

A side of an equilateral triangle is $20$ cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

Solution. The diagram shows triangles nested inside one another, each formed by joining the mid-points of the sides of the one outside it. Joining mid-points halves every side, so the side, and hence the perimeter, of each new triangle is half that of the previous one. The first triangle has perimeter $3 \times 20 = 60$ cm, so the perimeters form a G.P. with first term $60$ and ratio $\frac12$.

The original triangle is not itself inscribed in anything, so the sixth inscribed triangle comes after six halvings:

$$60 \times \left(\tfrac12\right)^6 = \frac{60}{64} = \frac{15}{16}\text{ cm}.$$

The printed key says $\frac{15}{8}$ cm. That is $60 \times \left(\frac12\right)^5$, the sixth triangle in the list when the original is counted as the first (it is the fifth inscribed triangle). The question’s wording points to $\frac{15}{16}$ cm; if your teacher counts the outer triangle as the first, the expected answer is $\frac{15}{8}$ cm. Whichever way you count, state which triangle you are using.

Question 8

In a potato race $20$ potatoes are placed in a line at intervals of $4$ metres with the first potato $24$ metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Solution. The potatoes are $24, 28, 32, \dots$ metres from the start, an A.P. with $a = 24$, $d = 4$, $n = 20$. The last one is at $24 + 19 \times 4 = 100$ m. For each potato he runs out and back, which is twice its distance, so the total is $2 \times (\text{sum of the distances})$:

$$S_{20} = \frac{20}{2}(24 + 100) = 1240, \qquad \text{total} = 2 \times 1240 = 2480 \text{ m}.$$

Question 9

In a cricket tournament $16$ school teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?

Solution. The 16 prizes form an A.P. Listed from last place upward, $a = 275$ and $n = 16$ with common difference $d$ unknown, and the sum is $8000$:

$$\frac{16}{2}[2 \times 275 + 15d] = 8000 \;\Rightarrow\; 550 + 15d = 1000 \;\Rightarrow\; d = 30.$$

First place is the 16th term in this order: $275 + 15 \times 30 = 725$.

Question 10

If $a_1, a_2, a_3, \dots, a_n$ are in A.P., where $a_i > 0$ for all $i$, show that

$$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \dots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$

Solution. Rationalise each term. Since consecutive terms differ by $d$, $a_{k+1} – a_k = d$, so

$$\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} – \sqrt{a_k}}{a_{k+1} – a_k} = \frac{\sqrt{a_{k+1}} – \sqrt{a_k}}{d}.$$

Adding for $k = 1$ to $n-1$, the middle square roots cancel in pairs (the sum telescopes):

$$\text{LHS} = \frac{\sqrt{a_n} – \sqrt{a_1}}{d}.$$

Also $a_n – a_1 = (n-1)d$, so $d = \dfrac{a_n – a_1}{n-1}$, and

$$\text{LHS} = \frac{(n-1)(\sqrt{a_n} – \sqrt{a_1})}{a_n – a_1} = \frac{(n-1)(\sqrt{a_n} – \sqrt{a_1})}{(\sqrt{a_n} – \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$

Question 11

Find the sum of the series $(3^3 – 2^3) + (5^3 – 4^3) + (7^3 – 6^3) + \dots$ to (i) $n$ terms (ii) $10$ terms.

Solution. The $k$th term is $(2k+1)^3 – (2k)^3$. Expanding $(2k+1)^3 = 8k^3 + 12k^2 + 6k + 1$ and subtracting $8k^3$ leaves $12k^2 + 6k + 1$. Now sum over $k = 1$ to $n$ using $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:

$$S_n = 12\cdot\frac{n(n+1)(2n+1)}{6} + 6\cdot\frac{n(n+1)}{2} + n = 2n(n+1)(2n+1) + 3n(n+1) + n = 4n^3 + 9n^2 + 6n.$$

For $n = 1$ this gives $19 = 3^3 – 2^3$, which checks. For $n = 10$: $4000 + 900 + 60 = 4960$.

Question 12

Find the $r$th term of an A.P. the sum of whose first $n$ terms is $2n + 3n^2$. [Hint: $a_n = S_n – S_{n-1}$.]

Solution. The $r$th term is the difference of two consecutive sums:

$$T_r = S_r – S_{r-1} = (2r + 3r^2) – [2(r-1) + 3(r-1)^2] = 2 + 3(2r – 1) = 6r – 1.$$

This also works for $r = 1$, since $T_1 = 5 = S_1$.

Question 13

If $A$ is the arithmetic mean and $G_1$, $G_2$ be two geometric means between any two numbers, then prove that $2A = \dfrac{G_1^2}{G_2} + \dfrac{G_2^2}{G_1}$.

Solution. Let the two numbers be $a$ and $b$. Then $a, G_1, G_2, b$ is a G.P. with ratio $r$, where $b = ar^3$, $G_1 = ar$, $G_2 = ar^2$, and $A = \frac{a+b}{2}$. Compute each fraction:

$$\frac{G_1^2}{G_2} = \frac{a^2r^2}{ar^2} = a, \qquad \frac{G_2^2}{G_1} = \frac{a^2r^4}{ar} = ar^3 = b.$$

So the right side is $a + b = 2A$.

Question 14

If $\theta_1, \theta_2, \theta_3, \dots, \theta_n$ are in A.P., whose common difference is $d$, show that

$$\sec\theta_1\sec\theta_2 + \sec\theta_2\sec\theta_3 + \dots + \sec\theta_{n-1}\sec\theta_n = \frac{\tan\theta_n – \tan\theta_1}{\sin d}.$$

Solution. Because $\theta_{k+1} – \theta_k = d$,

$$\sin d = \sin(\theta_{k+1} – \theta_k) = \sin\theta_{k+1}\cos\theta_k – \cos\theta_{k+1}\sin\theta_k.$$

Divide by $\cos\theta_k\cos\theta_{k+1}$:

$$\frac{\sin d}{\cos\theta_k\cos\theta_{k+1}} = \tan\theta_{k+1} – \tan\theta_k \;\Rightarrow\; \sec\theta_k\sec\theta_{k+1} = \frac{\tan\theta_{k+1} – \tan\theta_k}{\sin d}.$$

Adding for $k = 1$ to $n-1$, the tangents cancel in pairs and only the ends remain:

$$\text{LHS} = \frac{\tan\theta_n – \tan\theta_1}{\sin d}.$$

Question 15

If the sum of $p$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$, show that the sum of $p+q$ terms is $-(p+q)$. Also, find the sum of the first $p-q$ terms ($p > q$).

Solution. We are given $\frac p2[2a + (p-1)d] = q$ and $\frac q2[2a + (q-1)d] = p$. Subtract the second from the first:

$$a(p – q) + \frac d2\left[(p^2 – p) – (q^2 – q)\right] = q – p.$$

Since $p^2 – q^2 – (p – q) = (p-q)(p+q-1)$, divide by $p – q$ (non-zero):

$$a + \frac d2(p + q – 1) = -1 \;\Rightarrow\; 2a + (p+q-1)d = -2.$$

Then $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] = \frac{p+q}{2}(-2) = -(p+q)$.

For the second part we need $d$. From $2a = -2 – (p+q-1)d$, the first given condition becomes

$$\frac p2\left[-2 – (p+q-1)d + (p-1)d\right] = q \;\Rightarrow\; \frac p2(-2 – qd) = q \;\Rightarrow\; d = \frac{-2(p+q)}{pq}.$$

Now

$$S_{p-q} = \frac{p-q}{2}\left[2a + (p-q-1)d\right] = \frac{p-q}{2}\left[-2 – (p+q-1)d + (p-q-1)d\right] = \frac{p-q}{2}(-2 – 2qd).$$

With the value of $d$, $-2 – 2qd = -2 + \dfrac{4(p+q)}{p} = \dfrac{2p + 4q}{p}$, so

$$S_{p-q} = \frac{(p-q)(p+2q)}{p}.$$

(Sympy confirms both results by solving for $a$ and $d$ directly.)

Question 16

If the $p$th, $q$th and $r$th terms of an A.P. and a G.P. are both $a$, $b$ and $c$ respectively, show that $a^{b-c}\cdot b^{c-a}\cdot c^{a-b} = 1$.

Solution. Let the G.P. have first term $A$ and ratio $R$, so $a = AR^{p-1}$, $b = AR^{q-1}$, $c = AR^{r-1}$. Then

$$a^{b-c}\,b^{c-a}\,c^{a-b} = A^{(b-c)+(c-a)+(a-b)}\,R^{(p-1)(b-c) + (q-1)(c-a) + (r-1)(a-b)}.$$

The exponent of $A$ is $0$. In the exponent of $R$ the “$-1$” parts add to $-[(b-c)+(c-a)+(a-b)] = 0$, leaving $p(b-c) + q(c-a) + r(a-b)$. Now use the A.P., with common difference $D$: $b – c = (q – r)D$, $c – a = (r – p)D$, $a – b = (p – q)D$. Then

$$p(q-r)D + q(r-p)D + r(p-q)D = D\,(pq – pr + qr – qp + rp – rq) = 0.$$

So the whole expression is $A^0 R^0 = 1$.

Common mistakes

• Counting the wrong terms in Question 1. “The next $q$ terms” start at term $p+1$, so the sum is $S_{p+q} – S_p$, not $S_q$. Students who forget that the first $p$ terms are already used drop the subtraction.
• Choosing the wrong starting point in Question 9. Prizes grow as places improve, but the data are given from the last place. Start the A.P. at Rs 275 and take the 16th term; starting from the top gives a negative $d$ and an easy sign slip.
• Forgetting the return trip in Question 8. The contestant brings potatoes back one at a time, so every distance is doubled. $1240$ m is only the one-way total.
• Ignoring the negative root in Question 5. $n^2 + 4n – 192 = 0$ has the roots $12$ and $-16$. Days cannot be negative, and saying so is part of the answer.
• Dropping the shift in $T_r = S_r – S_{r-1}$ (Question 12). Substituting $r-1$ into $3n^2$ gives $3(r-1)^2$, not $3r^2 – 1$.
• Dividing by $p – q$ without noting it is non-zero (Question 15). The step is valid because $p \neq q$; the question itself assumes $p > q$ for the second part.
• Mis-counting the triangles in Question 7. The outer triangle is not inscribed. Decide whether you are counting inscribed triangles or triangles in the list, and then use $\left(\frac12\right)^{\text{halvings}}$.

Practise next

• Exemplar Objective, Fill in the Blanks and True or False Q17–36, the rest of the unit, on A.P. and G.P. properties.
• NCERT textbook Exercise 8.1 for general terms and sums of series.
• NCERT textbook Exercise 8.2 for G.P. questions like Questions 4 and 13.
• NCERT textbook Miscellaneous Exercise on Chapter 8 for special series like Question 11.