NCERT Exemplar Class 11 Maths solutions Chapter 9 Sequence and Series

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:04 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 9 has been assembled very carefully by our experts. These solutions make students aware of various concepts and methods associated with sequence and series through the easiest ways accepted by CBSE. A sequence in Mathematics has the same sense as it is in English, which denotes the arrangement of numbers in some specific order so that every number could be identified independently. NCERT Exemplar Class 11 Maths chapter 9 solutions cover a variety of topics linked with sequence and series. Class 11 Maths NCERT exemplar solutions chapter 9 also help 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.

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NCERT Exemplar Class 11 Maths Solutions Chapter 9- Excercise: 1.3
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NCERT Exemplar Class 11 Maths Solutions Chapter 9- Excercise: 1.3

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 $\frac{-a(p+q) q}{p-1}$ .
[Hint: Required sum $S\textsubscript{p+q} - S\textsubscript{p}$ ]

Answer:

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

Replacing the value of d in equation (i)

$\\ \text {required sum}=\frac{p+q}{2}\left(2 a+(p+q-1)\left(-\frac{2 a}{p-1}\right)\right. \\ =\frac{p+q}{2}\left(2 a-\frac{2 a p+2 a q-2 a}{p-1}\right) \\ =\frac{p+q}{2}(2 a)\left(1-\frac{p-1}{p-1}-\frac{q}{p-1}\right) \\ \text {On simplifying,} \\ =a(p+q)\left(-\frac{q}{p-1}\right)=-\frac{a(p+q) q}{p-1}$

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?

Answer:

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

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

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

$t_{n}=a+ \left( n-1 \right) d$

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

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

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

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

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

Question:4

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

Answer:

The nth term is given by $t_{n}=ar^{n-1}$
The pth term is given by $t_{p}=ar^{p-1}$
$q=ar^{p-1}$
$\frac{q}{r^{p}}=\frac{a}{r}$
$t_{q}=ar^{q-1}$
$p=ar^{q-1}$
$\\ \frac{p}{r^{q}}=\frac{a}{r}\\\\ \frac{p}{r^{q}}=\frac{q}{r^{p}}\\\\ r^{p-q}=\frac{q}{p}\\$
$r= \left( \frac{q}{p} \right) ^{\frac{1}{p-q}}$
$t_{p+q}=a \left( r \right) ^{p+q-1}$
$t_{p+q}= \left( ar^{p-1} \right) r^{q} \\\\ t_{p+q}=qr^{q} \\\\ t_{p+q}=q \left( \left( \frac{q}{p} \right) ^{\frac{1}{p-q}} \right) ^{q} \\\\$
$t_{p+q}=q \left( \frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}} \right) = \left( \frac{q^{1+\frac{q}{p-q}}}{p^{\frac{q}{p-q}}} \right) = \left( \frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}} \right) = \left( \frac{q^{p}}{p^{q}} \right) ^{\frac{1}{p-q}} \\\\$

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

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

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

Since 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 interir angles of a polygon having 'n' sides is given by $\left( n-2 \right)\ast 180^{\circ}$
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 side increases by 1, the sum of interior angle increases by $180^{\circ}$
So,for sum of angles of polygon with 21 sides $= \left( n-2 \right) \ast180^{\circ}~ \\\\$
$\\ = \left( 21-2 \right) \ast 180^{\circ}=19\ast 180^{\circ}~ \\\\ =3420^{\circ} \\\\$

Question:7

A side of an equilateral triangle is 20cm 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.

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

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

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 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?

Answer:

It is given that at 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
$\\S_{n}= \left( \frac{n}{2} \right) \left( 2a+ \left( n-1 \right) d \right) \\\\ S_{20}= \left( \frac{20}{2} \right) \left( 2 \left( 24 \right) + \left( 20-1 \right) 4 \right) \\\\ S_{20}=10 \left( 48+76 \right) \\\\ S_{20}=1240m \\\\$

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 first place be Rs. A and ‘d’ be the difference 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 total prize given is of 8000 rs hence

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

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

Question:10

If $a\textsubscript{1}, a\textsubscript{2}, a\textsubscript{3}, $ \ldots $ , a\textsubscript{n}$ are in A.P. $a\textsubscript{i} > 0$ for all i show that $\frac{1}{\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{2}}}+\frac{1}{\sqrt{\mathrm{a}_{2}}+\sqrt{\mathrm{a}_{3}}}+\cdots+\frac{1}{\sqrt{\mathrm{a}_{\mathrm{n}-1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}}=\frac{\mathrm{n}-1}{\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}}$

Answer:

$LHS=\frac{1}{\sqrt[]{a_{1}}+\sqrt[]{a_{2}}}+\frac{1}{\sqrt[]{a_{2}}+\sqrt[]{a_{3}}}+ \ldots +\frac{1}{\sqrt[]{a_{n-1}}+\sqrt[]{a_{n}}} \\\\$
$\begin{aligned} &\text {Multiplying the first term by } \frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{\sqrt{a_{1}}-\sqrt{a_{2}}}, \text { the second term by } \frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{\sqrt{a_{2}}-\sqrt{a_{3}}}\\ &\text {and so on rationalising each term} \end{aligned}$

$LHS= \left( \frac{1}{\sqrt[]{a_{1}}+\sqrt[]{a_{2}}} \right) \left( \frac{\sqrt[]{a_{1}}-\sqrt[]{a_{2}}}{\sqrt[]{a_{1}}-\sqrt[]{a_{2}}} \right) + \left( \frac{1}{\sqrt[]{a_{2}}+\sqrt[]{a_{3}}} \right) \left( \frac{\sqrt[]{a_{2}}-\sqrt[]{a_{3}}}{\sqrt[]{a_{2}}-\sqrt[]{a_{3}}} \right) + \ldots + \left( \frac{1}{\sqrt[]{a_{n-1}}+\sqrt[]{a_{n}}} \right) \left( \frac{\sqrt[]{a_{n-1}}-\sqrt[]{a_{n}}}{\sqrt[]{a_{n-1}}-\sqrt[]{a_{n}}} \right) \\\\$
$\\= \left( \frac{\sqrt[]{a_{1}}-\sqrt[]{a_{2}}}{a_{1}-a_{2}} \right) + \left( \frac{\sqrt[]{a_{2}}-\sqrt[]{a_{3}}}{a_{2}-a_{3}} \right) + \ldots + \left( \frac{\sqrt[]{a_{n-1}}-\sqrt[]{a_{n}}}{a_{n-1}-a_{n}} \right) \\\\ = \left( \frac{\sqrt[]{a_{1}}-\sqrt[]{a_{2}}}{-d} \right) + \left( \frac{\sqrt[]{a_{2}}-\sqrt[]{a_{3}}}{-d} \right) + \ldots + \left( \frac{\sqrt[]{a_{n-1}}-\sqrt[]{a_{n}}}{-d} \right) \\\\$

$\\ =-\frac{1}{d} \left( \sqrt[]{a_{1}}-\sqrt[]{a_{2}}+\sqrt[]{a_{2}}-\sqrt[]{a_{3}}+ \ldots +\sqrt[]{a_{n-1}}-\sqrt[]{a_{n}} \right) \\\\ =-\frac{1}{d} \left( \sqrt[]{a_{1}}-\sqrt[]{a_{n}} \right) ~ \\\\$

Rationalising again,

$\\ =-\frac{1}{d} \left( \sqrt[]{a_{1}}-\sqrt[]{a_{n}} \right) \left( \frac{ \left( \sqrt[]{a_{1}}+\sqrt[]{a_{n}} \right) }{ \left( \sqrt[]{a_{1}}+\sqrt[]{a_{n}} \right) } \right) \\\\ =-\frac{1}{d}\frac{ \left( a_{1}-a_{n} \right) }{ \left( \sqrt[]{a_{1}}+\sqrt[]{a_{n}} \right) }=-\frac{1}{d}\frac{ \left( - \left( n-1 \right) d \right) }{ \left( \sqrt[]{a_{1}}+\sqrt[]{a_{n}} \right) }=\frac{ \left( n-1 \right) }{ \left( \sqrt[]{a_{1}}+\sqrt[]{a_{n}} \right) } \\\\$

Question:11

Find the sum of the series $(3\textsuperscript{3} - 2\textsuperscript{3}) + (5\textsuperscript{3} - 4\textsuperscript{3}) + (7\textsuperscript{3} - 6\textsuperscript{3}) + \ldots$ to (i) n terms
(ii) 10 terms

Answer:

Generalising the series in terms of i
$\\ S= \sum _{i=1}^{n} \left[ \left( 2i+1 \right) ^{3}- \left( 2i \right) ^{3} \right] \\\\ = \sum _{i=1}^{n} \left[ \left( 2i+1-2i \right) \left( \left( 2i+1 \right) ^{2}+ \left( 2i+1 \right) \left( 2i \right) + \left( 2i \right) ^{2} \right) \right] \\\\ = \sum _{i=1}^{n} \left[ 4i^{2}+4i+1+4i^{2}+2i+4i^{2} \right] = \sum _{i=1}^{n} \left[ 12i^{2}+6i+1 \right] \\\\$
$\\ =12 \left[ \frac{n \left( n+1 \right) \left( 2n+1 \right) }{6} \right] +6 \left[ \frac{n \left( n+1 \right) }{2} \right] + \left[ n \right] \\\\ =2 \left[ 2n^{3}+3n^{2}+n \right] +3 \left[ n^{2}+n \right] + \left[ n \right] \\\\$
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 $2n + 3n^2$ .
$[Hint: a\textsubscript{n} = S\textsubscript{n} - S\textsubscript{n-1}]$

Answer:

Sum of first n terms is given by $S_n=2n + 3n^2$
The rth term is
$\\a_{r}=S_{r}-S_{r-1}= \left( 2r+3r^{2} \right) - \left( 2 \left( r-1 \right) +3 \left( r-1 \right) ^{2} \right) \\\\ =2r+3r^{2}- \left( 2r-2+3r^{2}-6r+3 \right) \\\\ =2r+3r^{2}- \left( 3r^{2}-4r+1 \right) \\\\ =6r-1 \\\\$

Question:13

If A is the arithmetic mean and $G_1$ , $G_2$ be two geometric means between any two numbers, then prove that
$2 \mathrm{A}=\frac{\mathrm{G}_{1}^{2}}{\mathrm{G}_{2}}+\frac{\mathrm{G}_{2}^{2}}{\mathrm{G}_{1}}$

Answer:

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

Question:14

$If \theta \textsubscript{1}, \theta \textsubscript{2}, \theta \textsubscript{3}, \ldots , \theta \textsubscript{n}$ are in A.P., whose common difference is d, show that

Answer:

Given that $\theta _{1}, \theta _{2}, \ldots \theta _{n}\text{~are in A.P and common difference is d. } \\\\$

$\\ \sec \theta _{1}\sec \theta _{2}+\sec \theta _{2}\sec \theta _{3}+ \ldots +\sec \theta _{n-1}\sec \theta _{n} \\\\ =\frac{1}{\cos \theta _{1}\cos \theta _{2}}+\frac{1}{\cos \theta _{2}\cos \theta _{3}}+ \ldots +\frac{1}{\cos \theta _{n-1}\cos \theta _{n}} \\\\ =\frac{\sin d}{\sin d} \left( \frac{1}{\cos \theta _{1}\cos \theta _{2}}+\frac{1}{\cos \theta _{2}\cos \theta _{3}}+ \ldots +\frac{1}{\cos \theta _{n-1}\cos \theta _{n}} \right) \\\\$
$\\ =\frac{1}{\sin d} \left( \frac{\sin \left( \theta _{2}- \theta _{1} \right) }{\cos \theta _{1}\cos \theta _{2}}+\frac{\sin \left( \theta _{3}- \theta _{2} \right) }{\cos \theta _{2}\cos \theta _{3}}+ \ldots +\frac{\sin \left( \theta _{n}- \theta _{n-1} \right) }{\cos \theta _{n-1}\cos \theta _{n}} \right) \\\\ =\frac{1}{\sin d} \left( \frac{\sin \theta _{2}\cos \theta _{1}-\cos \theta _{2}\sin \theta _{1}}{\cos \theta _{1}\cos \theta _{2}}+\frac{\sin \theta _{3}\cos \theta _{2}-\cos \theta _{3}\sin \theta _{2}}{\cos \theta _{2}\cos \theta _{3}}+ \ldots +\frac{\sin \theta _{n}\cos \theta _{n-1}-\cos \theta _{n}\sin \theta _{n-1}}{\cos \theta _{n-1}\cos \theta _{n}} \right) \\\\$
$\\=\frac{1}{\sin d} \left( \frac{\sin \theta _{2}\cos \theta _{1}}{\cos \theta _{1}\cos \theta _{2}}-\frac{\sin \theta _{1}\cos \theta _{2}}{\cos \theta _{1}\cos \theta _{2}}+\frac{\sin \theta _{3}\cos \theta _{2}}{\cos \theta _{2}\cos \theta _{3}}-\frac{\sin \theta _{2}\cos \theta _{3}}{\cos \theta _{2}\cos \theta _{3}}+ \ldots +\frac{\sin \theta _{n}\cos \theta _{n-1}}{\cos \theta _{n-1}\cos \theta _{n}}-\frac{\sin \theta _{n}\cos \theta _{n-1}}{\cos \theta _{n-1}\cos \theta _{n}} \right) \\\\ =\frac{1}{\sin d} \left( \tan \theta _{2}-\tan \theta _{1}+\tan \theta _{3}-\tan \theta _{2}+ \ldots +\tan \theta _{n}-\tan \theta _{n-1} \right) \\\\$

$\\ =\frac{ \left( \tan \theta _{n}-\tan \theta _{1} \right) }{\sin d} \\\\$

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 first p - q terms (p > q).

Answer:

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

$\\ S_{p}=q \\\\ q=\frac{p}{2} \left( 2a+ \left( p-1 \right) d \right) \\\\ \frac{2q}{p}=2a+ \left( p-1 \right) d \\\\ S_{q}=p \\\\ p=\frac{q}{2} \left( 2a+ \left( q-1 \right) d \right) \\\\$
$\\ \frac{2p}{q}=2a+ \left( q-1 \right) d \\\\ \frac{2p}{q}-\frac{2q}{p}= \left( q-1 \right) d- \left( p-1 \right) d \\\\ \frac{2p}{q}-\frac{2q}{p}= \left( q-p \right) d \\\\ d=\frac{2p^{2}-2q^{2}}{pq \left( q-p \right) }=-\frac{2 \left( p+q \right) }{pq} \\\\$
$\\ S_{p+q}=\frac{p+q}{2} \left( 2a+ \left( p+q-1 \right) d \right) \\\\ =\frac{p}{2} \left( 2a+ \left( p-1 \right) d+qd \right) +\frac{q}{2} \left( 2a+ \left( q-1 \right) d+pd \right) \\\\$
$\\ =\frac{p}{2} \left( 2a+ \left( p-1 \right) d \right) +\frac{pqd}{2}+\frac{q}{2} \left( 2a+ \left( q-1 \right) d \right) +\frac{pqd}{2} \\\\ =q+p+pqd \\\\$
$\\ =q+p+pq \left( -\frac{2 \left( p+q \right) }{pq} \right) =q+p-2p-2q \\\\ S_{p+q}=- \left( p+q \right) \\\\ S_{p-q}=\frac{p-q}{2} \left( 2a+ \left( p-q-1 \right) d \right) \\\\$
$\\ =\frac{p}{2} \left( 2a+ \left( p-1 \right) d \right) -\frac{pqd}{2}-\frac{q}{2} \left( 2a+ \left( p-1 \right) d \right) +\frac{q^{2}d}{2} \\\\ =q-\frac{pqd}{2}- \left( \frac{q}{p} \right) \left( \frac{p}{2} \right) \left( 2a+ \left( p-1 \right) d \right) +\frac{q^{2}d}{2} \\\\ =q-\frac{q^{2}}{p}-\frac{pqd}{2}+\frac{q^{2}d}{2} \\\\ =\frac{ \left( pq-q^{2} \right) }{p}+\frac{d \left( q^{2}-pq \right) }{2} \\\\ =\frac{pq-q^{2}}{p}-\frac{pq-q^{2}}{2} \left( -\frac{2 \left( p+q \right) }{pq} \right) \\\\$
$\\ =\frac{pq-q^{2}}{p}+ \left( pq-q^{2} \right) \left[ \frac{1}{q}+\frac{1}{p} \right] \\\\ =\frac{2 \left( pq-q^{2} \right) }{p}+\frac{pq-q^{2}}{q} \\\\ =\frac{2q \left( p-q \right) }{p}+p-q \\\\$

Question:16

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

Answer:

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

For AP

$\\ t_{p}=m+ \left( p-1 \right) d \\\\ a=m+ \left( p-1 \right) d \\\\ b=m+ \left( q-1 \right) d \\\\ c=m+ \left( r-1 \right) d \\\\$

For GP
$\\ t_{p}=Is^{p-1} \\\\ a=Is^{p-1} \\\\ b=Is^{q-1} \\\\ c=Is^{r-1} \\\\ b-c= \left( q-r \right) d \\\\$

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

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

Answer:

Given that $S_{n}=3n+2n^{2}$
We know that $a_{2}=S_{2}-S_{1}$

$\\ = \left( 3 \left( 2 \right) +2 \left( 2 \right) ^{2} \right) - \left( 3 \left( 1 \right) +2 \left( 1 \right) ^{2} \right) \\\\ =14-5 \\\\ =9 \\\\ a_{1}=S_{1}=3 \left( 1 \right) +2 \left( 1 \right) ^{2}=5 \\\\ d=a_{2}-a_{1}=9-5=4 \\\\$

Hence, 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 $T_{3}=4$

$\\ T_{n}=ar^{n-1} \\\\ ar^{2}=4 \\\\$
Product of first 5 terms

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

Hence, 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:

$\\ 9a_{9}=13a_{13} \\\\ 9 \left( a+ \left( 9-1 \right) d \right) =13 \left( a+ \left( 13-1 \right) d \right) \\\\ 9a+72d=13a+156d \\\\ -4a=84d \\\\ a=-21d \\\\ a_{22}=a+ \left( 22-1 \right) d \\\\ =-21d+21d=0 \\\\$

Hence, 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

$\\ 2y-x=3z-2y \\\\ x=4y-3z \\\\ \frac{y}{x}=\frac{z}{y} \\\\ y^{2}=xz \\\\ y^{2}= \left( 4y-3z \right) z \\\\ y^{2}-4yz+3z^{2}=0 \\\\ \left( 3z-y \right) \left( z-y \right) =0 \\\\ y=3z \ or\ y=z \\\\ r=\frac{y}{z}=\frac{1}{3} \\\\$

Hence, correct option is b.

Question:21

If in an A.P., Sn = qn² and Sm = qm², where S_r denotes the sum of r terms of the A.P., then S_q 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 first term ‘a’ and common difference ‘d’.

$\\ S_{n}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\\\ qn^{2}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\\\ 2qn=2a+ \left( n-1 \right) d \\\\ 2a=2qn- \left( n-1 \right) d \\\\ S_{m}=\frac{n}{2} \left[ 2a+ \left( m-1 \right) d \right] \\\\ qm^{2}=\frac{m}{2} \left[ 2a+ \left( m-1 \right) d \right] \\\\$

$\\2qm=2a+ \left( m-1 \right) d \\\\ 2a=2qm- \left( m-1 \right) d \\\\ 2qm- \left( m-1 \right) d=2qn- \left( n-1 \right) d \\\\ 2q \left( n-m \right) = \left( n-m \right) d \\\\ d=2q \\\\ 2a=2qn- \left( n-1 \right) \left( 2q \right) \\\\$

$\\a=q \\\\ S_{q}=\frac{q}{2} \left[ 2q+ \left( q-1 \right) 2q \right] =q^{3} \\\\$

Hence, correct option is c.

Question:22

Let $S_n$ denote the sum of the first n terms of an A.P. If $S\textsubscript{2n} = 3S\textsubscript{n}$ then $S\textsubscript{3n} : S\textsubscript{n}$ is equal to
A. 4
B. 6
C. 8
D. 10

Answer:

Given $S_{2n}=3S_{n} \\\\$

$\\ S_{n}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\\\ S_{2n}=\frac{2n}{2} \left[ 2a+ \left( 2n-1 \right) d \right] \\\\ S_{2n}=3S_{n} \\\\ n \left[ 2a+ \left( 2n-1 \right) d \right] =\frac{3n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\\\ 4a+4nd-2d=6a+3nd-3d \\\\ 2a=nd+d \\\\$

$\\ S_{3n}=\frac{3n}{2} \left[ 2a+ \left( 3n-1 \right) d \right] =\frac{3n}{2} \left[ \left( n+1 \right) d+ \left( 3n-1 \right) d \right] =6n^{2}d \\\\ S_{n}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] =\frac{n}{2} \left[ \left( n+1 \right) d+ \left( n-1 \right) d \right] =n^{2}d \\\\ \frac{S_{3n}}{S_{n}}=\frac{6n^{2}d}{n^{2}d}=6 \\\\$

Hence, the correct option is b.

Question:23

The minimum value of $4\textsuperscript{x} + 4\textsuperscript{1-x}, x \in R$ is
A. 2
B. 4
C. 1
D. 0

Answer:

The AM-GM inequality states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list

$\\ \frac{x+y}{2} \geq \sqrt {xy} \\\\ \frac{4^{x}+4^{1-x}}{2} \geq \sqrt { \left( 4^{x} \right) \left( 4^{1-x} \right) } \\\\ \frac{4^{x}+4^{1-x}}{2} \geq \sqrt {4} \\\\ 4^{x}+4^{1-x} \geq 4 \\\\$

Hence, correct option is b.

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 $\sum_{r=1}^{n} \frac{S_{r}}{s_{r}}$ equals.
A. $\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{6} \\$

B. $\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\$
C. $\frac{\mathrm{n}^{2}+3 \mathrm{n}+2}{2}$
D. None of these

Answer:

Sum of cubes of first n natural numbers $S_{n}= \sum _{i=1}^{n}i^{3}= \left( \frac{n \left( n+1 \right) }{2} \right) ^{2} \\\\$

Sum of first n natural numbers $s_{n}= \sum _{i=1}^{n}i=\frac{n \left( n+1 \right) }{2} \\\\$

$\\ \sum _{i=1}^{n}\frac{S_{n}}{s_{n}}= \sum _{i=1}^{n}\frac{n^{2}}{2}+\frac{n}{2} \\\\ =\frac{1}{2} \left[ \sum _{i=1}^{n}n^{2}+ \sum _{i=1}^{n}n \right] \\\\ =\frac{1}{2} \left[ \frac{n \left( n+1 \right) \left( 2n+1 \right) }{6}+\frac{n \left( n+1 \right) }{2} \right] \\\\ =\frac{1}{2} \left[ \frac{n \left( n+1 \right) }{2} \left( \frac{2n+1}{3}+1 \right) \right] \\\\$
$\\ =\frac{1}{2}\frac{n \left( n+1 \right) }{2}\frac{ \left( 2n+4 \right) }{3} \\\\ =\frac{n \left( n+1 \right) \left( n+2 \right) }{6} \\\\$

Hence, correct option is (a).

Question:25

If $t_n$ denotes the $n^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + ...$ then $t_{50}$ is
A. $49^2 - 1$
B. $49^2$
C. $50^2 + 1$
D. $49^2 + 2$

Answer:

$S_{n}=2+3+6+11+18+ \ldots +t_{50}$ (Using method of difference)
$\\ S_{n}=0+2+3+6+11+18+ \ldots +t_{50} \\\\ S_{n}-S_{n}= \left( 2-0 \right) + \left( 3-2 \right) + \left( 6-3 \right) + \left( 11-6 \right) + \left( 18-11 \right) + \ldots -t_{50} \\\\ t_{50}=2+ \left[ 1+3+5+7+ \ldots upto\: \: 49\: \: terms \right] \\\\ t_{50}=2+\frac{49}{2} \left[ 2+ \left( 49-1 \right) 2 \right] \\\\ t_{50}=2+49^{2} \\\\$
Hence, correct option is d.

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
A. 12 cm
B. 6 cm
C. 18 cm
D. 3 cm

Answer:

The volume and total S.A of a block are given
Let the length, breadth and height of a rectangular block be given by $\frac{a}{r},a,ar$ respectively.

$V=L*B *H=\frac{a}{r} \times a \times ar=a^{3} \\\\$
$\\216=a^{3} \\\\ a=6cm \\\\$
$S=2 \left[ L*B+B*H+H*L \right] =2 \left[ \frac{a}{r} \times a+\frac{a}{r} \times ar+a \times ar \right] =2 \left( \frac{a^{2}}{r}+a^{2}+a^{2}r \right) \\\\$
$\\ 252=2a^{2} \left( \frac{1}{r}+1+r \right) \\\\\frac{252}{72}=\frac{r^{2}+r+1}{r} \\\\ 2r^{2}-5r+2=0 \\\\ \left( 2r-1 \right) \left( r-2 \right) =0 \\\\ r=\frac{1}{2} \ or \ r=2 \\\\$

If a = 6 and r=2 then l,b and h are $\text{3 cm, 6 cm,12 cm}$ respectively.

If a=6 and r=1/2 then l,b and h are 12, 6 and 3 respectively.

In both the cases, the longest side is 12 units.

Hence, correct option is a.

Question:27

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

Answer:

It is given that a, b, c are in G.P
$\\ \frac{b}{a}=\frac{c}{b}=r \\\\ \frac{a-b}{b-c}=\frac{\frac{a-b}{b}}{\frac{b-c}{b}}=\frac{\frac{a}{b}-1}{1-\frac{c}{b}}=\frac{\frac{1}{r}-1}{1-r}=\frac{\frac{1-r}{r}}{1-r} \\\\ =\frac{1}{r}=\frac{a}{b}=\frac{b}{c} \\\\$

Question:28

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

Answer:

Taking first and last terms of
$a_{1}+a_{n}=a+a+ \left( n-1 \right) d=2a+ \left( n-1 \right) d$
Taking k-th and (n-k+1) -th term
$\\ a_{k}+a_{n-k+1}=a+ \left( k-1 \right) d+ \left( a+ \left( n-k+1-1 \right) d \right) \\\\ =2a+ \left( k-1+n-k+1-1 \right) d \\\\ =2a+ \left( n-1 \right) d \\\\ =a_{1}+a_{n} \\\\$

Question:29

Fill in the blanks The third term of a G.P. is 4, the product of the first five terms is ................

Answer:

The third term of a G.P is given by $T_{3}=4$

Let the common ratio be ‘r’

$\\T_{1}=\frac{4}{r^{2}} , T_{2}=\frac{4}{r} , ~T_{4}=4r , T_{5}=4r^{2} \\\\$

Product of first 5 terms $= \left( \frac{4}{r^{2}} \right) \left( \frac{4}{r} \right) \left( 4 \right) \left( 4r \right) \left( 4r^{2} \right) =4^{5}=1024 \\\\$

Question:30

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

Answer:

Let the terms of A.P be $a, \left( a+d \right) , \left( a+2d \right) , \ldots$
Let the terms of G.P be $a,ar,ar^{2}$

$\\ a+d=ar \\\\ a+2d=ar^{2} \\\\ r=\frac{a+2d}{a+d}=\frac{a+d}{a} \\\\ a^{2}+2ad=a^{2}+2ad+d^{2} \\\\ d^{2}=0 \\\\ r=\frac{a+2 \left( 0 \right) }{a+ \left( 0 \right) }=1 \\\\$

So the terms of A.P and G.P are same

Question:31

State whether statement in True or False. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.

Answer:

A progression is a sequence, but a sequence is not a progression because it does not follow a specific pattern.
Hence, the given statement is TRUE.

Question:32

State whether statement in True or False.
Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.

Answer:

Let us consider an A.P a, a+d, a+2d, a+3d
$\\a_{2}+a_{4}= \left( a+d \right) + \left( a+3d \right) =2 \left( a+2d \right) =2 \left( a_{3} \right) ~~~~ \\\\ ~a_{3}=\frac{a_{2}+a_{4}}{2}~ \\\\$
Hence, the given statement is TRUE.

Question:33

State whether statement in True or False. The sum or difference of two G.P.s, is again a G.P.

Answer:

Let the two G.P be $a,ar_{1},ar_{1}^{2},ar_{1}^{3}, \ldots$
And $b,br_{2},br_{2}^{2},br_{2}^{3}, \ldots \\\\$
The terms of sum of these two G.P are

$\\T_{1}= \left( b+a \right) \\\\ T_{2}= \left( br_{2}+ar_{1} \right) \\\\ T_{3}= \left( br_{2}^{2}+ar_{1}^{2} \right) \\\\ r^{'}=\frac{T_{3}}{T_{2}}=\frac{T_{2}}{T_{1}} \\\\$
$\\ \frac{T_{3}}{T_{2}}=\frac{br_{2}^{2}+ar_{1}^{2}}{br_{2}+ar_{1}}~and~\frac{T_{2}}{T_{1}}=\frac{br_{2}+ar_{1}}{b+a} \\\\ \frac{T_{3}}{T_{2}} \neq \frac{T_{2}}{T_{1}} \\\\$

The terms of difference of two G.P are

$\\ T_{1}= \left( b-a \right) \\\\ T_{2}= \left( br_{2}-ar_{1} \right) \\\\ T_{3}= \left( br_{2}^{2}-ar_{1}^{2} \right) \\\\ r^{'}=\frac{T_{3}}{T_{2}}=\frac{T_{2}}{T_{1}} \\\\$
$\\ \frac{br_{2}^{2}-ar_{1}^{2}}{br_{2}-ar_{1}}=\frac{br_{2}-ar_{1}}{b-a} \\\\ \frac{T_{3}}{T_{2}} \neq \frac{T_{2}}{T_{1}} \\\\$

Hence, the difference and sum of 2 G.P is not a G.P because common ratio is not same.

Question:34

State whether statement in True or False. If the sum of n terms of a sequence is quadratic expression then it always represents an A.P.

Answer:

$S_{n}=an^{2}+bn+c \\\\$
For n=1 $S_{1}=a+b+c \\\\$

$\\ S_{2}=4a+2b+c \\\\ S_{3}=9a+3b+c \\\\ a_{1}=S_{1}=a+b+c \\\\ a_{2}=S_{2}-S_{1}=3a+b \\\\ a_{3}=S_{3}-S_{2}=5a+b \\\\$
$\\d=a_{2}-a_{1}=2a-c \\\\ d=a_{3}-a_{2}=2a \\\\ a_{3}-a_{2} \neq a_{2}-a_{1} \\\\$

So, it does not represent an AP because common difference is not same.

Question:35

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

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

Answer:

$\left( a \right) \text{ Given 4, 1, }\frac{1}{4},\frac{1}{16}\text{~~ Here,}\frac{a_{2}}{a_{1}}=\frac{1}{4}~,\frac{a_{3}}{a_{2}}=\frac{\frac{1}{4}}{1}=\frac{1}{4}\text{~ and similarly}\frac{a_{4}}{a_{3}}=\frac{1}{16} *\frac{4}{1}=\frac{1}{4}~~ \\\\$

Hence, the given numbers is in G.P with common ratio $\frac{1}{4}~~ \\\\$
$\\~ \left( a \right) - \left( iii \right) ~~~~ \\\\ \left( b \right) \text{~Given~2, 3, 5, 7 } \\\\ ~Here,~a_{2}-a_{1}=3-2=1~~~ a_{3}-a_{2}=5-3=2~~ a_{4}-a_{3}=7-5=2~~ \\\\$
$~a_{2}-a_{1} \neq a_{3}-a_{2}~~ \\\\$

Hence, it is not in AP

Now, we will check the ratio $\frac{a_{2}}{a_{1}}=\frac{3}{2},~\frac{a_{3}}{a_{2}}=\frac{5}{7}~~ \\\\$

$\\ ~So,\frac{a_{2}}{a_{1}} \neq \frac{a_{3}}{a_{2}}\text{~~Thus,~it~is not a G.P } \\\\ ~ \left( b \right) - \left( ii \right) ~~ \\\\ ~ \left( c \right) ~13,8,~3, -2, -7 Here, a_{2}-a_{1}=8-13=-5 \\\\$

$\\~a_{3}-a_{2}=3-8=~-5 \\\\ ~a_{4}-a_{3}=-2-3=-5~ \\\\ ~~~~a_{2}-a_{1}=a_{3}-a_{2}\text{ Hence, it is an A.P } \\\\ ~~~ \left( c \right) - \left( i \right) \\\\$

Question:36

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

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

Answer:

Sum of ‘n’ natural numbers $\sum _{i=1}^{n}i= \left( \frac{n \left( n+1 \right) }{2} \right) \\\\$
Sum of cube of ‘n’ natural numbers $\sum _{i=1}^{n}i^{3}= \left( \frac{n \left( n+1 \right) }{2} \right) ^{2} \\\\$
Sum of square of ‘n’ natural numbers $\sum _{i=1}^{n}i^{2}= \left( \frac{n \left( n+1 \right) \left( 2n+1 \right) }{6} \right) \\\\$
Sum of twice times ‘n’ natural numbers $\sum _{i=1}^{n}2i=n \left( n+1 \right) \\\\$

So, (a) – (iii), (b)- (i), (c)- (ii), (d)- (iv)

What will the students learn in NCERT Exemplar Class 11 Maths Solutions Chapter 9?

The students will learn different concepts of sequence and series that will help them in other mathematical disciplines such as differential equations and their analysis.
NCERT Exemplar Class 11 Maths solutions Chapter 9 also helps in developing the analytical thinking, critical thinking and reasoning of the learners, which is helpful in competitive exams along with real-life situations.
The concept of sequence and series are also helpful in various different subjects for higher education, along with its use in subjects like physics.
The concept also gives an insight into its application in various different fields like astronomy, physics, chemistry, biology, finance, architecture, medicine and much more in every other aspect of science and mathematics.

NCERT Exemplar Class 11 Maths Solutions Chapter-Wise

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 4 Principle of Mathematical Induction

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 10 Straight lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability

Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 9 Sequence and Series

Some of the important topics for students to review are as follows:

· The students should learn about different sequences, including arithmetic and geometric progressions, its properties along with its importance in mathematics.

· NCERT Exemplar Class 11 Maths chapter 9 solutions cover the sum of elements of any sequence referred to as series whether finite or infinite and its representation through the compact form of sigma notation.

· NCERT Exemplar Class 11 Maths solutions chapter 9 also includes the study of arithmetic and the geometric mean of sequences.

· The students must practice the numerical and solutions of this chapter from the exam point of view.

· NCERT Exemplar solutions for Class 11 Maths chapter 9 also derives the relation between the arithmetic mean and the geometric mean of any sequence and its formulas.

Check Chapter-Wise NCERT Solutions of Book

Chapter-1	Sets
Chapter-2	Relations and Functions
Chapter-3	Trigonometric Functions
Chapter-4	Principle of Mathematical Induction
Chapter-5	Complex Numbers and Quadratic equations
Chapter-6	Linear Inequalities
Chapter-7	Permutation and Combinations
Chapter-8	Binomial Theorem
Chapter-9	Sequences and Series
Chapter-10	Straight Lines
Chapter-11	Conic Section
Chapter-12	Introduction to Three Dimensional Geometry
Chapter-13	Limits and Derivatives
Chapter-14	Mathematical Reasoning
Chapter-15	Statistics
Chapter-16	Probability

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Physics Solutions

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Frequently Asked Questions (FAQs)

1. What are some of the important topics of this chapter?

Some of the important topics of NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series are arithmetic progression, arithmetic mean, geometric progression, geometric mean, and the relationship between geometric mean and arithmetic mean.

2. Are these solutions useful from an exam point of view?

Yes, NCERT exemplar Class 11 Maths solutions chapter 9 are very important for any type of exam as they provide a better understanding and explanation of different types of concept through easier methods.

3. Where can I download the solutions for this chapter?

NCERT Exemplar solutions for Class 11 Maths Chapter 9 can be easily downloaded in PDF format by using the webpage pdf download online tools.

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NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Limits and Derivatives

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Maths solutions Chapter 9 Sequence and Series

NCERT Exemplar Class 11 Maths Solutions Chapter 9- Excercise: 1.3

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NCERT Exemplar Class 11 Solutions

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