NCERT Exemplar Class 11 Maths solutions Chapter 9 Sequence and Series

NCERT Exemplar Class 11 Maths Solutions Chapter 9 has been assembled very carefully by our experts. In our daily lives, we often come across patterns—like the increasing number of chairs in each row of an auditorium, or savings added to a piggy bank every month. These patterns are perfect examples of sequences and series . A sequence is simply a list of numbers arranged in a particular order, while a series is the sum of the terms of a sequence. In Class 11, you’ll learn about arithmetic and geometric progressions, how to find terms, and how to calculate their sums. These concepts are not only useful in mathematics but also in finance, computer science, and real-world planning.

NCERT Exemplar Class 11 Maths chapter 9 solutions cover a variety of topics linked with sequences and series. Class 11 Maths NCERT exemplar solutions chapter 9 also helps with establishing relationships between different terms of sequence and series, followed by varied numerical problems and their solutions for practice and a better understanding of the concept.
Also, check - NCERT Class 11 Solutions Maths for other chapters.

Question 1: The first term of an A.P. is a, and the sum of the first p terms is zero, showing that the sum of its next q terms is $\frac{-a(p+q)q}{p-1}$. [Hint: Required sum $S_{\mathrm{p}+\mathrm{q}}-S_{\mathrm{p}}$ ]

Answer:

The sum of the first p terms and the first term is given. We have to find the next q terms. So, the total terms become p+q.
Hence, the sum of all terms minus the sum of the first p terms will give the sum of the next q terms. However, the sum of the first p terms is zero, so the sum of the next q terms will be the same as the sum of all terms.
Sum of n terms is given by $S_n=\left(\frac{n}{2}\right)(2a+(n-1)d)$
$$\begin{gathered} \text{ required sum }=S_{p+q}-S_p=\frac{p+q}{2}(2a+(p+q-1)d)\dots \dots (i) \\ {S}_p=\frac{p}{2}(2a+(p-1)d \\ 0=2a+(p-1)d \\ d=-\frac{2a}{p-1} \end{gathered}$$

Replacing the value of d in equation (i)

$$\begin{gathered} \text{required sum}=\frac{p+q}{2}(2a+(p+q-1)(-\frac{2a}{p-1}) \\ =\frac{p+q}{2}(2a-\frac{2ap+2aq-2a}{p-1}) \\ =\frac{p+q}{2}(2a)(1-\frac{p-1}{p-1}-\frac{q}{p-1}) \\ \text{On simplifying,} \\ =a(p+q)(-\frac{q}{p-1})=-\frac{a(p+q)q}{p-1} \end{gathered}$$

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?

Answer:

Total amount saved in 20 years is $S_{20}=66000$
Let the amount saved in the first year be ‘a.’
Since he increases by 200 Rs every year, d=200
Sum formula of A.P is $S_n=\left(\frac{n}{2}\right)(2a+(n-1)d)$

$S_{20}=\left(\frac{20}{2}\right)(2a+(20-1)200)$
$66000=10(2a+3800)$
$a=1400$

Question 3: A man accepts a position with an initial salary of Rs 5200 per month. It is understood that 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?

Answer:

The man’s salary in the first month is Rs. 5200 and increases every month by Rs. 320. So, we have the value of ‘a’ and ‘d’.

$t_n=a+(n-1)d$

Salary in 10 ^th month is given by $t_{10}=5200+(10-1)320=8080$

For total earnings in the first year, the first 12 terms of the sequence are added.

$S_n=\left(\frac{n}{2}\right)(2a+(n-1)d)$

$S_{12}=\left(\frac{12}{2}\right)(2\left(5200\right)+(12-1)320)$

$S_{12}=6(10400+11(320)=83520$

Question:4 If the $p^{\text{th }}$ and $q^{\text{th }}$ terms of a G.P. are q and p respectively, show that its $(p+q{)}^{\text{th }}$ term is ${\left(\frac{q^p}{p^4}\right)}^{\frac{1}{p-8}}$.

Answer:

The nth term is given by $t_n=a{r}^{n-1}$
The pth term is given by $t_p=a{r}^{p-1}$
$q=a{r}^{p-1}$
$\frac{q}{r^p}=\frac{a}{r}$
$t_q=a{r}^{q-1}$
$p=a{r}^{q-1}$
$$\begin{gathered} \frac{p}{r^q}=\frac{a}{r} \\ \frac{p}{r^q}=\frac{q}{r^p} \\ {r}^{p-q}=\frac{q}{p} \end{gathered}$$
$r={\left(\frac{q}{p}\right)}^{\frac{1}{p-q}}$
$t_{p+q}=a{\left(r\right)}^{p+q-1}$
$$\begin{gathered} t_{p+q}=\left(a{r}^{p-1}\right)r^q \\ {t}_{p+q}=q{r}^q \\ {t}_{p+q}=q{\left({\left(\frac{q}{p}\right)}^{\frac{1}{p-q}}\right)}^q \end{gathered}$$
$t_{p+q}=q\left(\frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}}\right)=\left(\frac{q^{1+\frac{q}{p-q}}}{p^{\frac{q}{p-q}}}\right)=\left(\frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}}\right)={\left(\frac{q^p}{p^q}\right)}^{\frac{1}{p-q}}$.

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?

Answer:

The first day he made 5 frames and 2 more frames each passing day. So,
a=5 and d=2
$S_n=\left(\frac{n}{2}\right)(2a+(n-1)d)$

Let us assume that it takes ‘n’ days to build 192 window frames.

$$\begin{gathered} {S}_n=\left(\frac{n}{2}\right)(10+(n-1)2) \\ 192=\left(\frac{n}{2}\right)(8+2n) \\ 384=2n^2+8n \\ {n}^2+4n-192=0 \\ (n-12)(n+16)=0 \end{gathered}$$

Since the number of days can’t be negative, n = 12.

Question 6:

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

Answer:

The sum of interior angles of a polygon having 'n' sides is given by $(n-2)\ast {180}^{\circ }$
By this formula sum of angles with 3 sides, 4 sides, 5 sides, and 6 sides is ${180}^{\circ },{360}^{\circ },{540}^{\circ },{720}^{\circ }$ respectively. As the number of sides increases by 1, the sum of the interior angles increases by ${180}^{\circ }$
So, for the sum of angles of a polygon with 21 sides $=(n-2)\times {180}^{\circ }$
$$\begin{gathered} =(21-2)\times {180}^{\circ }=19\times {180}^{\circ } \\ ={3420}^{\circ } \end{gathered}$$

Question:7

A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the midpoints 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.

Answer:

Let AB=BC=AC=20cm
Let D, E, and F be midpoints of AC, CB, and AB, which are joined to form an equilateral triangle DEF. So, CD=CE=10cm. Triangle CDE is equilateral. Hence, DE = 10 cm. Similarly, GH = 5 cm
The series of sides of an equilateral triangle will be 20,10,5…..
The series is a G.P. with first term 20 and common ratio =1/2
$t_n=20{\left(\frac{1}{2}\right)}^{n-1}$

For the perimeter of the 6 ^th triangle, we first have to find the side of the 6 ^th triangle

$t_6=\frac{20}{2^{6-1}}=\frac{20}{2^5}=\frac{5}{8}cm$

The perimeter would be thrice the length of its side.

Perimeter of triangle= $3\left(\frac{5}{8}\right)=\frac{15}{8}cm$

Question:8

In a potato race, 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 meters 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?

Answer:

It is given that at the start he has to run 24m to get the first potato, then 28 m as the next potato is 4m away, and so on.
Hence, the sequence of its running will be 24, 28, 32…..
To get potatoes from the starting point he has to run 24+28+32+….. up to 20 terms.
$$\begin{gathered} {S}_n=\left(\frac{n}{2}\right)(2a+(n-1)d) \\ {S}_{20}=\left(\frac{20}{2}\right)(2\left(24\right)+(20-1)4) \\ {S}_{20}=10(48+76) \\ {S}_{20}=1240m \end{gathered}$$

Also, he has to get the potato back to the starting point, hence the total distance will be twice.
Total Distance ran=2×1240=2480m

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?

Answer:

Let the amount received by the first place be Rs. A and ‘d’ are the differences in the amount. The last team receives Rs. 275. As there are 16 teams and all teams are given prizes hence the sequence will have 16 terms because there are 16 teams.

As the total prize given is Rs 8000, hence

$a+a-d+a-2d....+275=8000$

$n=16$
$S_{16}=8000$
$S_n=\left(\frac{n}{2}\right)(2a+(n-1)(-d))$
$8000=\left(\frac{16}{2}\right)(2a+(16-1)(-d))=8(2a-15d)$
$$\begin{gathered} 1000=2a-15d \\ {t}_{16}=275 \end{gathered}$$
$t_{16}=a+(16-1)(-d)$
$275=a-15d$
$1000-275=(2a-15d)-(a-15d)$
$a=725$

Question:10 If $a_1,a_2,a_3,\dots ,a_{\mathrm{n}}$ are in A.P., where $a_{\mathrm{i}}>0$ for all i , show that $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$

Answer:

$LHS=\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$
$\begin{array}{rl}& \text{Multiplying the first term by }\frac{\sqrt{a_1}-\sqrt{a_2}}{\sqrt{a_1}-\sqrt{a_2}},\text{ the second term by }\frac{\sqrt{a_2}-\sqrt{a_3}}{\sqrt{a_2}-\sqrt{a_3}}\\ & \text{and so on rationalising each term}\end{array}$

$LHS=\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}\right)\left(\frac{\sqrt{a_1}-\sqrt{a_2}}{\sqrt{a_1}-\sqrt{a_2}}\right)+\left(\frac{1}{\sqrt{a_2}+\sqrt{a_3}}\right)\left(\frac{\sqrt{a_2}-\sqrt{a_3}}{\sqrt{a_2}-\sqrt{a_3}}\right)+\dots +\left(\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)\left(\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{\sqrt{a_{n-1}}-\sqrt{a_n}}\right)$
$$\begin{gathered} =\left(\frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}\right)+\left(\frac{\sqrt{a_2}-\sqrt{a_3}}{a_2-a_3}\right)+\dots +\left(\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-a_n}\right) \\ =\left(\frac{\sqrt{a_1}-\sqrt{a_2}}{-d}\right)+\left(\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}\right)+\dots +\left(\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}\right) \end{gathered}$$

$$\begin{gathered} =-\frac{1}{d}(\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}-\sqrt{a_3}+\dots +\sqrt{a_{n-1}}-\sqrt{a_n}) \\ =-\frac{1}{d}(\sqrt{a_1}-\sqrt{a_n}) \end{gathered}$$

Rationalising again,

$$\begin{gathered} =-\frac{1}{d}(\sqrt{a_1}-\sqrt{a_n})\left(\frac{(\sqrt{a_1}+\sqrt{a_n})}{(\sqrt{a_1}+\sqrt{a_n})}\right) \\ =-\frac{1}{d}\frac{(a_1-a_n)}{(\sqrt{a_1}+\sqrt{a_n})}=-\frac{1}{d}\frac{(-(n-1)d)}{(\sqrt{a_1}+\sqrt{a_n})}=\frac{(n-1)}{(\sqrt{a_1}+\sqrt{a_n})} \end{gathered}$$

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

Answer:

Generalizing the series in terms of i
$S=\sum _{i=1}^n[{(2i+1)}^3-{\left(2i\right)}^3]$
$=\sum _{i=1}^n\left[(2i+1-2i)({(2i+1)}^2+(2i+1)\left(2i\right)+{\left(2i\right)}^2)\right]$
$=\sum _{i=1}^n[4i^2+4i+1+4i^2+2i+4i^2]=\sum _{i=1}^n[12i^2+6i+1]$
$$\begin{gathered} =12\left[\frac{n(n+1)(2n+1)}{6}\right]+6\left[\frac{n(n+1)}{2}\right]+\left[n\right] \\ =2[2n^3+3n^2+n]+3[n^2+n]+\left[n\right] \end{gathered}$$
Sum up to n terms $=4n^3+9n^2+6n$
$\text{Sum up to 10 terms }{S}_{10}=4{\left(10\right)}^3+9{\left(10\right)}^2+6\left(10\right)=4960$

Question:12 Find the rth term of an A.P. sum of whose first n terms is $2n+3n^2$.

$[$ Hint : $a_{\mathrm{n}}=S_{\mathrm{n}}-S_{\mathrm{n}-1}]$

Answer:

The sum of the first n terms is given by $S_n=2n+3n^2$
The rth term is
$$\begin{gathered} {a}_r=S_r-S_{r-1}=(2r+3r^2)-(2(r-1)+3{(r-1)}^2) \\ =2r+3r^2-(2r-2+3r^2-6r+3) \\ =2r+3r^2-(3r^2-4r+1) \\ =6r-1 \end{gathered}$$

Question:13 If A is the arithmetic mean and $G_1$, and $G_2$ are two geometric means between any two numbers, then prove that.
$2\mathrm{A}=\frac{{\mathrm{G}}_1^2}{{\mathrm{G}}_2}+\frac{{\mathrm{G}}_2^2}{{\mathrm{G}}_1}$

Answer:

It is given that A is the arithmetic mean and $G_1$, and $G_2$ are two geometric means between any two numbers.
$$\begin{gathered} A=\frac{a+b}{2} \\ G=\sqrt{ab} \\ {G}_1=\sqrt{a{G}_2} \\ {G}_2=\sqrt{G_{1}b} \\ {G}_1^2=a{G}_2 \end{gathered}$$
$$\begin{gathered} {G}_1^2=a\sqrt{G_{1}b} \\ {G}_1^4=a^2{G}_{1}b \\ {G}_1^3=a^{2}b \\ {G}_1=a^{\frac{2}{3}}{b}^{\frac{1}{3}} \end{gathered}$$
$$\begin{gathered} {G}_2={(a^{\frac{2}{3}}{b}^{\frac{1}{3}}\times b)}^{\frac{1}{2}}=a^{\frac{1}{3}}{b}^{\frac{2}{3}} \\ \frac{G_1^2}{G_2}=\left(\frac{{\left(a^{\frac{2}{3}}{b}^{\frac{1}{3}}\right)}^2}{a^{\frac{1}{3}}{b}^{\frac{2}{3}}}\right)=a \\ \frac{G_2^2}{G_1}=\left(\frac{{\left(b^{\frac{2}{3}}{a}^{\frac{1}{3}}\right)}^2}{b^{\frac{1}{3}}{a}^{\frac{2}{3}}}\right)=b \\ \frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=a+b=2A \end{gathered}$$

Question:14 If ${\theta }_1,{\theta }_2,{\theta }_3,\dots ,{\theta }_{\mathrm{n}}$ are in A.P., whose common difference is d, show that $\sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sec {\theta }_3+\dots +\sec {\theta }_{\mathrm{n}-1}\sec {\theta }_{\mathrm{n}}$ $=\frac{\tan {\theta }_{\mathrm{n}}-\tan {\theta }_1}{\sin \mathrm{d}}$

Answer:

Given that ${\theta }_1,{\theta }_2,\dots {\theta }_n\text{~are in A.P and common difference is d. }$

$$\begin{gathered} \sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sec {\theta }_3+\dots +\sec {\theta }_{n-1}\sec {\theta }_n \\ =\frac{1}{\cos {\theta }_1\cos {\theta }_2}+\frac{1}{\cos {\theta }_2\cos {\theta }_3}+\dots +\frac{1}{\cos {\theta }_{n-1}\cos {\theta }_n} \\ =\frac{\sin d}{\sin d}(\frac{1}{\cos {\theta }_1\cos {\theta }_2}+\frac{1}{\cos {\theta }_2\cos {\theta }_3}+\dots +\frac{1}{\cos {\theta }_{n-1}\cos {\theta }_n}) \end{gathered}$$
$$\begin{gathered} =\frac{1}{\sin d}(\frac{\sin ({\theta }_2-{\theta }_1)}{\cos {\theta }_1\cos {\theta }_2}+\frac{\sin ({\theta }_3-{\theta }_2)}{\cos {\theta }_2\cos {\theta }_3}+\dots +\frac{\sin ({\theta }_n-{\theta }_{n-1})}{\cos {\theta }_{n-1}\cos {\theta }_n}) \\ =\frac{1}{\sin d}(\frac{\sin {\theta }_2\cos {\theta }_1-\cos {\theta }_2\sin {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}+\frac{\sin {\theta }_3\cos {\theta }_2-\cos {\theta }_3\sin {\theta }_2}{\cos {\theta }_2\cos {\theta }_3}+\dots +\frac{\sin {\theta }_n\cos {\theta }_{n-1}-\cos {\theta }_n\sin {\theta }_{n-1}}{\cos {\theta }_{n-1}\cos {\theta }_n}) \end{gathered}$$
$=\frac{1}{\sin d}(\frac{\sin {\theta }_2\cos {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}-\frac{\sin {\theta }_1\cos {\theta }_2}{\cos {\theta }_1\cos {\theta }_2}+\frac{\sin {\theta }_3\cos {\theta }_2}{\cos {\theta }_2\cos {\theta }_3}-\frac{\sin {\theta }_2\cos {\theta }_3}{\cos {\theta }_2\cos {\theta }_3}+\dots +\frac{\sin {\theta }_n\cos {\theta }_{n-1}}{\cos {\theta }_{n-1}\cos {\theta }_n}-\frac{\sin {\theta }_n\cos {\theta }_{n-1}}{\cos {\theta }_{n-1}\cos {\theta }_n})$
$=\frac{1}{\sin d}(\tan {\theta }_2-\tan {\theta }_1+\tan {\theta }_3-\tan {\theta }_2+\dots +\tan {\theta }_n-\tan {\theta }_{n-1})$

$=\frac{(\tan {\theta }_n-\tan {\theta }_1)}{\sin d}$

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).

Answer:

The sum of n terms is given by
$S_n=\left(\frac{n}{2}\right)(2a+(n-1)d)$
Where ‘a’ is the first term and ’d’ is the common difference

$$\begin{gathered} {S}_p=q \\ q=\frac{p}{2}(2a+(p-1)d) \\ \frac{2q}{p}=2a+(p-1)d \\ {S}_q=p \\ p=\frac{q}{2}(2a+(q-1)d) \end{gathered}$$
$$\begin{gathered} \frac{2p}{q}=2a+(q-1)d \\ \frac{2p}{q}-\frac{2q}{p}=(q-1)d-(p-1)d \\ \frac{2p}{q}-\frac{2q}{p}=(q-p)d \\ d=\frac{2p^2-2q^2}{pq(q-p)}=-\frac{2(p+q)}{pq} \end{gathered}$$
$$\begin{gathered} {S}_{p+q}=\frac{p+q}{2}(2a+(p+q-1)d) \\ =\frac{p}{2}(2a+(p-1)d+qd)+\frac{q}{2}(2a+(q-1)d+pd) \end{gathered}$$
$$\begin{gathered} =\frac{p}{2}(2a+(p-1)d)+\frac{pqd}{2}+\frac{q}{2}(2a+(q-1)d)+\frac{pqd}{2} \\ =q+p+pqd \end{gathered}$$
$$\begin{gathered} =q+p+pq(-\frac{2(p+q)}{pq})=q+p-2p-2q \\ {S}_{p+q}=-(p+q) \\ {S}_{p-q}=\frac{p-q}{2}(2a+(p-q-1)d) \end{gathered}$$
$=\frac{p}{2}(2a+(p-1)d)-\frac{pqd}{2}-\frac{q}{2}(2a+(p-1)d)+\frac{q^{2}d}{2}$
$=q-\frac{pqd}{2}-\left(\frac{q}{p}\right)\left(\frac{p}{2}\right)(2a+(p-1)d)+\frac{q^{2}d}{2}$
$=q-\frac{q^2}{p}-\frac{pqd}{2}+\frac{q^{2}d}{2}$
$=\frac{(pq-q^2)}{p}+\frac{d(q^2-pq)}{2}$
$=\frac{pq-q^2}{p}-\frac{pq-q^2}{2}(-\frac{2(p+q)}{pq})$
$$\begin{gathered} =\frac{pq-q^2}{p}+(pq-q^2)[\frac{1}{q}+\frac{1}{p}] \\ =\frac{2(pq-q^2)}{p}+\frac{pq-q^2}{q} \\ =\frac{2q(p-q)}{p}+p-q \end{gathered}$$

Question:16 If $p^{th},q^{\text{th }}$, and $r^{\text{th }}$ terms of an A.P. and G.P. are both $\mathbf{a},\mathbf{b}$ and $\mathbf{c}$ respectively, show that $a^{\mathrm{b}-\mathrm{c}}\cdot b^{\mathrm{c}-\mathrm{a}}\cdot c^{\mathrm{a}-\mathrm{b}}=1$

Answer:

Let the first term of AP be m and common difference as d. Let the first term of GP be I and the common ratio be s.

For AP

$$\begin{gathered} {t}_p=m+(p-1)d \\ a=m+(p-1)d \\ b=m+(q-1)d \\ c=m+(r-1)d \end{gathered}$$

For GP
$$\begin{gathered} {t}_p=I{s}^{p-1} \\ a=I{s}^{p-1} \\ b=I{s}^{q-1} \\ c=I{s}^{r-1} \\ b-c=(q-r)d \end{gathered}$$
$$\begin{gathered} c-a=(r-p)d \\ a-b=(p-q)d \\ LHS=a^{b-c}{b}^{c-a}{c}^{a-b} \\ ={\left(I{s}^{p-1}\right)}^{(q-r)d}{\left(I{s}^{q-1}\right)}^{(r-p)d}{\left(I{s}^{r-1}\right)}^{(p-q)d} \\ =I^{(q-r+r-p+p-q)d}{s}^{[(pq-pr-q+r)+(qr-qp-r+p)+(rp-rq-p+q)]d} \\ =I^0{s}^0=1 \end{gathered}$$

Question:17 If the sum of $\mathit{n}$ terms of an A.P. is given by $S_n=3n+2n^2$, then the common difference of the A.P. is

(A) 3 (B) 2 (C) 6 (D) 4

Answer:

Given that $S_n=3n+2n^2$
We know that $a_2=S_2-S_1$

$$\begin{gathered} =(3\left(2\right)+2{\left(2\right)}^2)-(3\left(1\right)+2{\left(1\right)}^2) \\ =14-5 \\ =9 \\ {a}_1=S_1=3\left(1\right)+2{\left(1\right)}^2=5 \\ d=a_2-a_1=9-5=4 \end{gathered}$$

Hence, the correct option is d.

Question:18 The third term of G.P. is 4. The product of its first 5 terms is (A) 43 (B) 44 (C) 45 (D) None of these

Answer:

The third term of G.P is given $T_3=4$

$$\begin{gathered} {T}_n=a{r}^{n-1} \\ a{r}^2=4 \end{gathered}$$
Product of the first 5 terms

$$\begin{gathered} =\left(a\right)\left(ar\right)\left(a{r}^2\right)\left(a{r}^3\right)\left(a{r}^4\right) \\ =\left(a^5{r}^{10}\right)={\left(a{r}^2\right)}^5 \\ =4^5=1024 \end{gathered}$$

Hence, the correct option is c.

Question:19

If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is A. 0 B. 22 C. 220 D. 198

Answer:

$$\begin{gathered} 9a_9=13a_{13} \\ 9(a+(9-1)d)=13(a+(13-1)d) \\ 9a+72d=13a+156d \\ -4a=84d \\ a=-21d \\ {a}_{22}=a+(22-1)d \\ =-21d+21d=0 \end{gathered}$$

Hence, the correct option is a.

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 A. 3 B. 1/3 C. 2 D. 1/2

Answer:

It is given that x, 2y, and 3z are in A.P

$$\begin{gathered} 2y-x=3z-2y \\ x=4y-3z \\ \frac{y}{x}=\frac{z}{y} \\ {y}^2=xz \\ {y}^2=(4y-3z)z \\ {y}^2-4yz+3z^2=0 \\ (3z-y)(z-y)=0 \\ y=3zory=z \\ r=\frac{y}{z}=\frac{1}{3} \end{gathered}$$

Hence, the correct option is b.

Question:21

If in an A.P., Sn = qn2 and Sm = qm2, where Sr denotes the sum of r terms of the A.P., then Sq equals A. $\frac{q^3}{2}$ B. mnq C. $q^3$ D. $(m+n)q^2$

Answer:

The given series is an A.P. with the first term ‘a’ and common difference ‘d’.

$$\begin{gathered} {S}_n=\frac{n}{2}[2a+(n-1)d] \\ q{n}^2=\frac{n}{2}[2a+(n-1)d] \\ 2qn=2a+(n-1)d \\ 2a=2qn-(n-1)d \\ {S}_m=\frac{n}{2}[2a+(m-1)d] \\ q{m}^2=\frac{m}{2}[2a+(m-1)d] \end{gathered}$$

$$\begin{gathered} 2qm=2a+(m-1)d \\ 2a=2qm-(m-1)d \\ 2qm-(m-1)d=2qn-(n-1)d \\ 2q(n-m)=(n-m)d \\ d=2q \\ 2a=2qn-(n-1)\left(2q\right) \end{gathered}$$

$$\begin{gathered} a=q \\ {S}_q=\frac{q}{2}[2q+(q-1)2q]=q^3 \end{gathered}$$

Hence, the correct option is c.

Question:22

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

A. 4

B. 6

C. 8

D. 10