NCERT Exemplar Solutions Class 11 Maths Sequences and Series — Objective, Fill in the Blanks, True or False and Matching Q17–36

NCERT Exemplar Problems, Class 11 Mathematics — Sequences and Series, Objective, Fill in the Blanks, True or False and Matching questions 17 to 36. All 20 questions solved.

This page has ten multiple-choice questions (17 to 26), three fill-in-the-blanks (27 to 29), five true-or-false statements (30 to 34) and two matching questions (35, 36). The multiple-choice questions mostly test one property each: the common difference from $S_n$ (17), the product of terms equidistant from the middle of a G.P. (18, 29), conditions that fix $a$ in terms of $d$ (19, 22), and AM-GM (23). The conceptual questions (30 to 35) ask whether you know the definitions precisely.

One answer in the printed key is wrong, the matching in Question 35. It is explained where it occurs.

Key insight. For an A.P. the sum is a quadratic in $n$ with no constant term, so $S_n = An^2 + Bn$ gives $d = 2A$. For a G.P. the product of terms equidistant from the middle is the square of the middle term. Most of these questions are over once you turn the given condition into a relation between $a$ and $d$ (or $a$ and $r$).

Question 17

If the sum of $n$ terms of an A.P. is given by $S_n = 3n + 2n^2$, then the common difference of the A.P. is

Solution. $T_n = S_n – S_{n-1} = 3 + 2(2n – 1) = 4n + 1$, so the terms are $5, 9, 13, \dots$ and the difference is $4$. Equivalently, the coefficient of $n^2$ in $S_n$ is $\frac d2$, so $d = 2 \times 2 = 4$.

Question 18

The third term of a G.P. is $4$. The product of its first $5$ terms is

Solution. Write the five terms as $\frac{a}{r^2}, \frac ar, a, ar, ar^2$ with the middle term $a = T_3 = 4$. Pairing equidistant terms, $\frac{a}{r^2}\cdot ar^2 = a^2$ and $\frac ar\cdot ar = a^2$, so the product is $a^2 \cdot a^2 \cdot a = a^5 = 4^5$.

Question 19

If $9$ times the $9$th term of an A.P. is equal to $13$ times the $13$th term, then the $22$nd term of the A.P. is

Solution. $9(a + 8d) = 13(a + 12d)$ gives $9a + 72d = 13a + 156d$, so $4a = -84d$ and $a = -21d$. Then $T_{22} = a + 21d = -21d + 21d = 0$.

Question 20

If $x$, $2y$, $3z$ are in A.P., where the distinct numbers $x$, $y$, $z$ are in G.P., then the common ratio of the G.P. is

Solution. Let $y = xr$ and $z = xr^2$. The A.P. condition $2(2y) = x + 3z$ becomes $4xr = x + 3xr^2$, that is $3r^2 – 4r + 1 = 0$, so $r = 1$ or $r = \frac13$. The numbers are distinct, which rules out $r = 1$, leaving $r = \frac13$.

Question 21

If in an A.P., $S_n = qn^2$ and $S_m = qm^2$, where $S_r$ denotes the sum of $r$ terms of the A.P., then $S_q$ equals

Solution. For an A.P., $S_r = Ar^2 + Br$ for constants $A$ and $B$. The two conditions give $(A – q)n + B = 0$ and $(A – q)m + B = 0$. Subtracting, $(A – q)(n – m) = 0$, and for $m \neq n$ this forces $A = q$ and then $B = 0$. So $S_r = qr^2$ for every $r$, and $S_q = q \cdot q^2 = q^3$.

Question 22

Let $S_n$ denote the sum of the first $n$ terms of an A.P. If $S_{2n} = 3S_n$, then $S_{3n} : S_n$ is equal to

Solution. $S_{2n} = n[2a + (2n-1)d]$ and $3S_n = \frac{3n}{2}[2a + (n-1)d]$. Equating and simplifying:

$$4a + 2(2n-1)d = 6a + 3(n-1)d \;\Rightarrow\; (n+1)d = 2a.$$

Then $2a + (3n-1)d = (n+1)d + (3n-1)d = 4nd$, so $S_{3n} = \frac{3n}{2}\cdot 4nd = 6n^2d$, while $2a + (n-1)d = 2nd$ gives $S_n = n^2d$. The ratio is $6$.

Question 23

The minimum value of $4^x + 4^{1-x}$, $x \in \mathbb R$, is

Solution. Both $4^x$ and $4^{1-x}$ are positive, so AM-GM applies: $4^x + 4^{1-x} \ge 2\sqrt{4^x\cdot4^{1-x}} = 2\sqrt4 = 4$. Equality needs $4^x = 4^{1-x}$, that is $x = \frac12$, which is allowed, so the minimum is attained and equals $4$.

Question 24

Let $S_n$ denote the sum of the cubes of the first $n$ natural numbers and $s_n$ denote the sum of the first $n$ natural numbers. Then $\displaystyle\sum_{r=1}^{n}\frac{S_r}{s_r}$ equals

Solution. $S_r = \left[\frac{r(r+1)}{2}\right]^2$ and $s_r = \frac{r(r+1)}{2}$, so $\frac{S_r}{s_r} = \frac{r(r+1)}{2}$. Then

$$\sum_{r=1}^n \frac{r(r+1)}{2} = \frac12\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)}{2}\cdot\frac{2n+4}{6} = \frac{n(n+1)(n+2)}{6}.$$

Check with $n = 2$: $1 + 3 = 4$ and $\frac{2\cdot3\cdot4}{6} = 4$.

Question 25

If $t_n$ denotes the $n$th term of the series $2 + 3 + 6 + 11 + 18 + \dots$, then $t_{50}$ is

Solution. The differences between consecutive terms are $1, 3, 5, 7, \dots$, odd numbers, which sum to a perfect square. So $t_n = 2 + (1 + 3 + \dots + (2n-3)) = 2 + (n-1)^2$. Check: $t_2 = 3$, $t_3 = 6$, $t_4 = 11$, $t_5 = 18$. Therefore $t_{50} = 49^2 + 2$.

Question 26

The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is $216$ cm$^3$ and the total surface area is $252$ cm$^2$. The length of the longest edge is

Solution. Take the edges as $\frac ar, a, ar$. The volume is $a^3 = 216$, so $a = 6$. The total surface area is $2\left(\frac{a^2}{r} + a^2 + a^2r\right) = 72\left(r + \frac1r + 1\right) = 252$, so $r + \frac1r = \frac52$, giving $r = 2$ or $\frac12$. Either way the edges are $3, 6, 12$ cm, and the longest is $12$ cm. (Option (B), $6$ cm, is the middle edge.)

Question 27

Fill in the blanks. For $a$, $b$, $c$ to be in G.P. the value of $\dfrac{a-b}{b-c}$ is equal to ____.

Solution. In a G.P., $b = ar$ and $c = ar^2$, so $\dfrac{a-b}{b-c} = \dfrac{a(1-r)}{ar(1-r)} = \dfrac1r = \dfrac ab$. Since $\frac ab = \frac bc$ in a G.P., $\frac bc$ is the same thing, which is why the key prints “$\frac ab$ or $\frac bc$”.

Question 28

The sum of terms equidistant from the beginning and end in an A.P. is equal to ____.

Solution. The $k$th term from the start is $a + (k-1)d$ and the $k$th from the end is $l – (k-1)d$, where $l$ is the last term. Their sum is $a + l$, whatever $k$ is. This is the fact behind the formula $S_n = \frac n2(a + l)$.

Question 29

The third term of a G.P. is $4$, the product of the first five terms is ____.

Solution. This is Question 18 in fill-in form: the product of five terms in G.P. is the fifth power of the middle term, $4^5 = 1024$.

Question 30

State whether the statement is True or False. Two sequences cannot be in both A.P. and G.P. together.

Solution. This is false. Read as “a sequence cannot be both an A.P. and a G.P.”, a counter-example is any non-zero constant sequence such as $5, 5, 5, \dots$: it has common difference $0$ and common ratio $1$. (A non-constant A.P. is never a G.P., which is why the statement feels true.)

Question 31

Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.

Solution. A progression has a fixed rule (constant difference or ratio), whereas a sequence can follow any rule, or none: $2, 3, 5, 7$ is a sequence of primes but not an A.P. or G.P. So the statement is true.

Question 32

Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.

Solution. If $T_k$ has a term $j$ places before it and another $j$ places after it, then $T_{k-j} + T_{k+j} = [a + (k-j-1)d] + [a + (k+j-1)d] = 2[a + (k-1)d] = 2T_k$. So $T_k = \frac12(T_{k-j} + T_{k+j})$. The book’s “(except first)” tacitly has an unending A.P. in mind; in a finite A.P. the last term has no partner either, so the statement holds for every term that has terms on both sides.

Question 33

The sum or difference of two G.P.s, is again a G.P.

Solution. Take $1, 2, 4, 8, \dots$ and $1, 3, 9, 27, \dots$. Their sum is $2, 5, 13, 35, \dots$, and $\frac52 \neq \frac{13}{5}$, so it is not a G.P. One counter-example is enough.

Question 34

If the sum of $n$ terms of a sequence is a quadratic expression then it always represents an A.P.

Solution. An A.P. has $S_n = An^2 + Bn$ with no constant term. Take $S_n = n^2 + 1$: $T_1 = 2$ but $T_2 = S_2 – S_1 = 3$ and $T_3 = S_3 – S_2 = 5$, so the differences are $1$ and $2$ and the sequence is not an A.P. The “always” is what makes the statement false.

Question 35

Match the questions given under Column I with their appropriate answers given under Column II.

Column I	Column II
(a) $4, 1, \frac14, \frac1{16}$	(i) A.P.
(b) $2, 3, 5, 7$	(ii) sequence
(c) $13, 8, 3, -2, -7$	(iii) G.P.

Solution. Test the ratio and the difference of consecutive terms.

• (a) The ratios $\frac14, \frac14, \frac14$ are constant, so it is a G.P.: (iii).
• (b) The differences $1, 2, 2$ and the ratios $\frac32, \frac53, \frac75$ are not constant, so it is only a sequence: (ii).
• (c) The differences $-5, -5, -5, -5$ are constant, so it is an A.P.: (i).

The printed key pairs (b) with (i) and (c) with (ii), which is the wrong way round. $2, 3, 5, 7$ cannot be an A.P. because $3 – 2 \neq 5 – 3$, and $13, 8, 3, -2, -7$ is plainly an A.P. with $d = -5$.

Question 36

Match the questions given under Column I with their appropriate answers given under Column II.

Column I	Column II
(a) $1^2 + 2^2 + 3^2 + \dots + n^2$	(i) $\left(\dfrac{n(n+1)}{2}\right)^2$
(b) $1^3 + 2^3 + 3^3 + \dots + n^3$	(ii) $n(n+1)$
(c) $2 + 4 + 6 + \dots + 2n$	(iii) $\dfrac{n(n+1)(2n+1)}{6}$
(d) $1 + 2 + 3 + \dots + n$	(iv) $\dfrac{n(n+1)}{2}$

Solution. These are the standard sums, and a quick check with $n = 2$ confirms each: $1 + 4 = 5 = \frac{2\cdot3\cdot5}{6}$; $1 + 8 = 9 = 3^2$; $2 + 4 = 6 = 2\cdot3$; $1 + 2 = 3 = \frac{2\cdot3}{2}$. The sum of the first $n$ even numbers is twice the sum of the first $n$ natural numbers, so (c) is $2 \cdot \frac{n(n+1)}{2} = n(n+1)$.

Common mistakes

• Reading the common difference off the wrong coefficient (Question 17). In $S_n = 3n + 2n^2$ the number $3$ is not the first term shift and $2$ is not $d$; $d$ is twice the coefficient of $n^2$. Students answer $2$ or $3$ because those are the numbers on the page.
• Keeping $r = 1$ in Question 20. The quadratic $3r^2 – 4r + 1 = 0$ has two roots, and the word “distinct” in the question is the only thing that removes $r = 1$.
• Stopping at $r$ in Question 26. The ratio $2$ or $\frac12$ is not the answer; the edges $3, 6, 12$ are. Picking $6$ cm (the middle edge) is the common slip.
• Forgetting AM-GM needs positive terms and an attainable equality (Question 23). Here $4^x$ and $4^{1-x}$ are always positive, and equality at $x = \frac12$ is allowed, so the bound $4$ is reached; a bound that cannot be reached is not a minimum.
• Thinking a constant sequence is neither (Question 30). $5, 5, 5, \dots$ has $d = 0$ and $r = 1$, so it is both an A.P. and a G.P.
• Believing “quadratic means A.P.” (Question 34). Only $S_n = An^2 + Bn$ with no constant term gives an A.P., and the constant term in $n^2 + 1$ breaks the pattern at the first term.
• Trusting the printed pairing in Question 35. The key swaps (b) and (c); test each list for a constant difference or ratio yourself.

Practise next

• Exemplar Short and Long Answer Q1–16, the proofs and word problems from this unit.
• NCERT textbook Exercise 8.1 for general terms and sums, the skill behind Questions 17, 19, 22 and 25.
• NCERT textbook Exercise 8.2 for G.P. questions like 18, 20 and 26.
• NCERT textbook Miscellaneous Exercise on Chapter 8 for special series like Questions 24 and 36.