Examples (Revised) - Chapter 9 - Sequences & Series - Ncert Solutions class 11 - Maths

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Chapter 8 - Sequences & Series | NCERT Solutions for Class 11 Maths

Example 1

Write the first three terms in each of the following sequences defined by the following:
(i) $a_n=2 n+5$,
(ii) $a_n=\frac{n-3}{4}$.

Solution

(i) Here $a_n=2 n+5$
Substituting $n=1,2,3$, we get
$
a_1=2(1)+5=7, a_2=9, a_3=11
$

Therefore, the required terms are 7,9 and 11 .
(ii) Here $a_n=\frac{n-3}{4}$. Thus, $a_1=\frac{1-3}{4}=-\frac{1}{2}, a_2=-\frac{1}{4}, a_3=0$

Hence, the first three terms are $-\frac{1}{2},-\frac{1}{4}$ and 0 .
Example 2

What is the $20^{\text {th }}$ term of the sequence defined by
$
a_n=(n-1)(2-n)(3+n) ?
$

Solution

Putting $n=20$, we obtain
$
\begin{aligned}
a_{20} & =(20-1)(2-20)(3+20) \\
& =19 \times(-18) \times(23)=-7866 .
\end{aligned}
$

Example 3

Let the sequence $a_n$ be defined as follows:
$
a_1=1, a_n=a_{n-1}+2 \text { for } n \geq 2 .
$

Find first five terms and write corresponding series.
Solution

We have
$
\begin{aligned}
& a_1=1, a_2=a_1+2=1+2=3, a_3=a_2+2=3+2=5, \\
& a_4=a_3+2=5+2=7, a_5=a_4+2=7+2=9 .
\end{aligned}
$

Hence, the first five terms of the sequence are $1,3,5,7$ and 9 . The corresponding series is $1+3+5+7+9+\ldots$

Example 4

Find the $10^{\text {th }}$ and $n^{\text {th }}$ terms of the G.P. $5,25,125, \ldots$.
Solution

Here $a=5$ and $r=5$. Thus, $a_{10}=5(5)^{10-1}=5(5)^9=5^{10}$ and $\quad a_n=a r^{n-1}=5(5)^{n-1}=5^n$.

Example 5

Which term of the G.P., $2,8,32, \ldots$ up to $n$ terms is 131072 ?
Solution

Let 131072 be the $n^{\text {th }}$ term of the given G.P. Here $a=2$ and $r=4$.
Therefore $\quad 131072=a_n=2(4)^{n-1}$ or $\quad 65536=4^{n-1}$
This gives $\quad 4^8=4^{n-1}$.
So that $n-1=8$, i.e., $n=9$. Hence, 131072 is the $9^{\text {th }}$ term of the G.P.
Example 6

In a G.P., the $3^{\text {rd }}$ term is 24 and the $6^{\text {th }}$ term is 192 . Find the $10^{\text {th }}$ term.
Solution

Here, $a_3=a r^2=24$
and $\quad a_6=a r^5=192$
Dividing (2) by (1), we get $r=2$. Substituting $r=2$ in (1), we get $a=6$.
Hence $a_{10}=6(2)^9=3072$.
Example 7

Find the sum of first $n$ terms and the sum of first 5 terms of the geometric series $1+\frac{2}{3}+\frac{4}{9}+\ldots$
Solution

Here $a=1$ and $r=\frac{2}{3}$. Therefore
$
\mathrm{S}_n=\frac{a\left(1-r^n\right)}{1-r}=\frac{\left[1-\left(\frac{2}{3}\right)^n\right]}{1-\frac{2}{3}}=3\left[1-\left(\frac{2}{3}\right)^n\right]
$

In particular, $\quad \mathrm{S}_5=3\left[1-\left(\frac{2}{3}\right)^5\right]=3 \times \frac{211}{243}=\frac{211}{81}$.

Example 8

How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the $\operatorname{sum} \frac{3069}{512} ?$

Solution

Let $n$ be the number of terms needed. Given that $a=3, r=\frac{1}{2}$ and $\mathrm{S}_n=\frac{3069}{512}$
Since
$
\mathrm{S}_n=\frac{a\left(1-r^n\right)}{1-r}
$

Therefore
$
\frac{3069}{512}=\frac{3\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^n}\right)
$
or
$
\begin{aligned}
\frac{3069}{3072} & =1-\frac{1}{2^n} \\
\frac{1}{2^n} & =1-\frac{3069}{3072}=\frac{3}{3072}=\frac{1}{1024}
\end{aligned}
$
or
$
2^n=1024=2^{10} \text {, which gives } n=10 \text {. }
$

Example 9

The sum of first three terms of a G.P. is $\frac{13}{12}$ and their product is -1 . Find the common ratio and the terms.

Solution

Let $\frac{a}{r}, a, a r$ be the first three terms of the G.P. Then
$
\frac{a}{r}+a r+a=\frac{13}{12}
$
and
$
\left(\frac{a}{r}\right)(a)(a r)=-1
$

From (2), we get $a^3=-1$, i.e., $a=-1$ (considering only real roots)
Substituting $a=-1$ in (1), we have
$
-\frac{1}{r}-1-r=\frac{13}{12} \text { or } 12 r^2+25 r+12=0 \text {. }
$

This is a quadratic in $r$, solving, we get $r=-\frac{3}{4}$ or $-\frac{4}{3}$.
Thus, the three terms of G.P. are $: \frac{4}{3},-1, \frac{3}{4}$ for $r=\frac{-3}{4}$ and $\frac{3}{4},-1, \frac{4}{3}$ for $r=\frac{-4}{3}$,
Example10

Find the sum of the sequence 7, 77, 777, 7777, ... to $n$ terms.
Solution

This is not a G.P., however, we can relate it to a G.P. by writing the terms as $\mathrm{S}_n=7+77+777+7777+\ldots$ to $n$ terms

$
\begin{aligned}
& =\frac{7}{9}[9+99+999+9999+\ldots \text { to } n \text { term }] \\
& =\frac{7}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\left(10^4-1\right)+\ldots n \text { terms }\right] \\
& =\frac{7}{9}\left[\left(10+10^2+10^3+\ldots n \text { terms }\right)-(1+1+1+\ldots n \text { terms })\right] \\
& =\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]=\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{9}-n\right] .
\end{aligned}
$

Example 11

A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.
Solution

Here $a=2, r=2$ and $n=10$
Using the sum formula $\mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1}$
We have
$
\mathrm{S}_{10}=2\left(2^{10}-1\right)=2046
$

Hence, the number of ancestors preceding the person is 2046 .

Example12

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.
Solution

Let $\mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3$ be three numbers between 1 and 256 such that $1, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3, 256$ is a G.P.

Therefore $\quad 256=r^4$ giving $r= \pm 4$ (Taking real roots only)
For $r=4$, we have $\mathrm{G}_1=a r=4, \mathrm{G}_2=a r^2=16, \mathrm{G}_3=a r^3=64$
Similarly, for $r=-4$, numbers are $-4,16$ and -64 .
Hence, we can insert $4,16,64$ between 1 and 256 so that the resulting sequences are in G.P.

Example 13

If A.M. and G.M. of two positive numbers $a$ and $b$ are 10 and 8 , respectively, find the numbers.

Solution

Given that
$
\text { A.M. }=\frac{a+b}{2}=10
$
and
$
\text { G.M. }=\sqrt{a b}=8
$

From (1) and (2), we get
$
\begin{aligned}
& a+b=20 \\
& a b=64
\end{aligned}
$

Putting the value of $a$ and $b$ from (3), (4) in the identity $(a-b)^2=(a+b)^2-4 a b$, we get
$
(a-b)^2=400-256=144
$
or
$
a-b= \pm 12
$

Solving (3) and (5), we obtain
$
a=4, b=16 \text { or } a=16, b=4
$

Thus, the numbers $a$ and $b$ are 4,16 or 16,4 respectively.