Exercise 5.5 - Chapter 5 Binomial Theorem Sequences and Series 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 5.5
Choose the correct or the most suitable answer:
Question 1.
The value of $2+4+6+\ldots+2 \mathrm{n}$ is ....
(a) $\frac{n(n-1)}{2}$
(b) $\frac{n(n+1)}{2}$
(c) $\frac{2 n(2 n+1)}{2}$
(d) $n(n+1)$
Solution:
(d) $n(n+1)$
Hint $.2+4+6+\ldots+2 n=2(1+2+3+\ldots n)=2\left[\frac{n(n+1)}{2}\right]=n(n+1)$
Question 2.
The coefficient of $x^6$ in $(2+2 x)^{10}$ is
(a) ${ }^{10} \mathrm{C}_6$
(b) $2^6$
(c) ${ }^{10} \mathrm{C}_6 2^6$
(d) ${ }^{10} \mathrm{C}_6 2^{10}$
Solution:
(d) ${ }^{10} \mathrm{C}_6 2^{10}$
Hint.
$\mathrm{t}_{\mathrm{r}+1}=2^{10}\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\right)$
To find coefficient of $\mathrm{x}_6$ put $\mathrm{r}=6$
$\therefore$ coefficient of $\mathrm{x}_6=210\left[{ }^{10} \mathrm{C}_6\right]$
Question 3.
The coefficient of $x^8 y^{12}$ in the expansion of $(2 x+3 y)^{20}$ is
(a) 0
(b) $2^8 3^{12}$
(c) $2^8 3^{12}+2^{12} 3^8$
(d) ${ }^{20} \mathrm{C}_8 2^8 3^{12}$
Solution:

(d) ${ }^{20} \mathrm{C}_8 2^8 3^{12}$
Hint. $t_{r+1}={ }^{20} \mathrm{C}_r(2 x)^{20-r}(3 y)^r={ }^{20} \mathrm{C}_r 2^{20-r}(x)^{20-r} 3^r y^r \quad\left[{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right]$
To find coefficient of $x^8$ put $20-r=8 \Rightarrow r=12$
To find coefficient of $y^{12}$ put $r=12 \therefore r=12$
Coefficient $={ }^{20} \mathrm{C}_{12}(2)^{20-12} 3^{12}={ }^{20} \mathrm{C}_{12} 2^8 3^{12}={ }^{20} \mathrm{C}_8 2^8 3^{12}$
Question 4.
If ${ }^n C_{10}>{ }^n C_r$ for all possible $r$, then a value of $n$ is
(a) 10
(b) 21
(c) 19
(d) 20
Solution:
(d) 20
Hint.
Out of ${ }^{10} \mathrm{C}_{10},{ }^{21} \mathrm{C}_{10},{ }^{19} \mathrm{C}_{10}$ and ${ }^{20} \mathrm{C}_{10},{ }^{20} \mathrm{C}_{10}$ is larger.
Question 5.
If $a$ is the arithmetic mean and $g$ is the geometric mean of two numbers, then
(a) $\mathrm{a} \leq \mathrm{g}$
(b) $a \geq g$
(c) $\mathrm{a}=\mathrm{g}$
(d) $a>g$
Solution:
(b) $\mathrm{a} \geq \mathrm{g}$
Hint. AM $\geq \mathrm{GM}$
$\therefore \mathrm{a} \geq \mathrm{g}$
Question 6.
If $\left(1+x^2\right)^2(1+x)^n=a_0+a_1 x+a_2 x^2+\ldots . .+x^{n+4}$ and if $a_0, a_1, a_2$ are in AP, then $n$ is .....
(a) 1
(b) 2
(c) 3
(d) 4
Solution:

(b or $\mathrm{c}) \mathrm{n}=2$ or 3
Hint. $\begin{aligned} &\left(1+x^2\right)^2(1+x)^n=\left(1+2 x^2+x^4\right)\left(1+n x+\frac{n(n-1)}{2} x^2 \ldots\right) \\ &=1+n x+\frac{n(n-1)}{2} x^2+2 x^2 \ldots=a_0+a_1 x+a_2 x^2+\ldots \text { (given) } \\ & \quad \text {} \\ & \Rightarrow a_0=1, a_1=n, a_2=\frac{n(n-1)}{2}+2(\text { i.e }) a_2=\frac{n^2-n+4}{2}\end{aligned}$
Given $a_0, a_1, a_2$ are in A.P.
$
\begin{aligned}
& 2 a_1=a_0+a_2 \Rightarrow 2 n=1+\frac{n^2-n+4}{2} \\
& \Rightarrow 2 n=\frac{n^2-n+6}{2} \Rightarrow n^2-n+6=4 n \\
& \text { (i.e) } n^2-5 n+6=0 \Rightarrow n=2 \text { or } 3
\end{aligned}
$
Question 7.
If $\mathrm{a}, 8, \mathrm{~b}$ are in A.P, a, 4, b are in G.P, if $\mathrm{a}, \mathrm{x}, \mathrm{b}$ are in HP then $\mathrm{x}$ is .....
(a) 2
(b) 1
(c) 4
(d) 16
Solution:
(a) 2
Hint: $a, 8, b$ are in AP $\Rightarrow \frac{a+b}{2}=8 \Rightarrow a+b=16$
$a, 4, b$ are in $\mathrm{GP} \Rightarrow a b=4^2=16$
Now $a, x, b$ are in HP $\Rightarrow x=\frac{2 a b}{a+b}=\frac{2(16)}{16}=2$

Question 8 .
The sequence $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2 \sqrt{2}} \ldots$ form an
(a) AP
(b) GP
(c) $\mathrm{HP}$
(d) AGP
Solution:
(c) $\mathrm{HP}$
Question 9.
The $\mathrm{HM}$ of two positive numbers whose $\mathrm{AM}$ and $\mathrm{GM}$ are 16,8 respectively is
(a) 10
(b) 6
(c) 5
(d) 4
Solution:
(d) 4
Hint
Let the two numbers be $\mathrm{a}$ and $\mathrm{b}$
$
\begin{aligned}
& \mathrm{AM}=\frac{a+b}{2}=16 \Rightarrow a+b=32 \\
& \mathrm{GM}=\sqrt{a b}=8 \Rightarrow a b=64 \\
& \mathrm{HM}=\frac{2 a b}{a+b}=\frac{2 \times 64}{32}=4
\end{aligned}
$

Question 10 .
If $S_n$ denotes the sum of $n$ terms of an $A P$ whose common difference is $d$, the value of $\mathrm{S}_n-2 \mathrm{~S}_{n-1}+S_{n-2}$ is $\ldots \ldots$
(a) $\mathrm{d}$
(b) $2 \mathrm{~d}$
(c) $4 \mathrm{~d}$
(d) $\mathrm{d}^2$
Solution:
(a) d
Hint. $\begin{aligned} S_n-2 S_{n-1}+S_{n-2} & =S_n-S_{n-1}-S_{n-1}+S_{n-2} \\ & =\left(S_n-S_{n-1}\right)-\left(S_{n-1}-S_{n-2}\right)\end{aligned}$
Now $S_n=t_1+t_2+\ldots t_n$
$S_{n-1}=t_1+t_2+\ldots t_{n-1}$ and $S_{n-2}=t_1+t_2+\ldots t_{n-2}$
So $S_n-S_{n-1}=t_n$ and $S_{n-1}-S_{n-2}=t_{n-1}$
Now $\left(S_n-S_{n-1}\right)-\left(S_{n-1}-S_{n-2}\right)=t_n-t_{n-1}$
$=[a+(n-1) d]-[a+(n-1-1) d]$
$=a+n d-d-a-n d+2 d=d$
Question 11.
The remainder when $38^{15}$ is divided by 13 is ......
(a) 12
(b) 1
(c) 11
(d) 5
Solution:

(a) 12
Hint
$
\begin{aligned}
& 38^{15}=(39-1)^{15}=39^{15}-15 \mathrm{C}_1 39^{14}(1)+15 \mathrm{C}_2(39)^{13}(1)^2-15 \mathrm{C}_3(39)^{12}(1)^3 \ldots .+15 \mathrm{C}_{14}(39)^1(1)- \\
& 15 \mathrm{C}_{15}(1)
\end{aligned}
$
Except -1 all other terms are divisible by 13 .
$\therefore$ When 1 is added to it the number is divisible by 13 . So the remainder is $13-1=12$.
Question 12 .
The $\mathrm{n}^{\text {th }}$ term of the sequence $1,2,4,7,11, \ldots \ldots$ is
(a) $n^3+3 n^2+2 n$
(b) $n^3-3 n^2+3 n$
(c) $\frac{n(n+1)(n+2)}{3}$
(d) $\frac{n^2-n+2}{2}$
Solution:
(d) $\frac{n^2-n+2}{2}$
Question 13.
The sum up to $n$ terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ldots$ is
(a) $\sqrt{2 n+1}$
(b) $\frac{\sqrt{2 n+1}}{2}$
(c) $\sqrt{2 n+1}-1$
(d) $\frac{\sqrt{2 n+1}-1}{2}$
Solution:
(d) $\frac{\sqrt{2 n+1-1}}{2}$
Hint. $\frac{1}{\sqrt{3}+1}=\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{3-1}=\frac{\sqrt{3}-\sqrt{1}}{2}$
$
\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{5-3}=\frac{\sqrt{5}-\sqrt{3}}{2}
$
So $\frac{1}{\sqrt{3}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots=\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\ldots .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n+1}}{2}\right)$ $=\frac{\sqrt{2 n+1}-1}{2}$

Question 14.
The $n^{\text {th }}$ term of the sequence $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}$ is
(a) $2^n-n-1$
(b) $1-2^{-n}$
(c) $2^{-n}+n-1$
(d) $2^{n-1}$
Solution:
$
\text { Hint. } \begin{aligned}
& t_1=\frac{1}{2}=1-\frac{1}{2} ; t_2=\frac{3}{4}=1-\frac{1}{4}=1-\frac{1}{2^2}, t_3=\frac{7}{8}=1-\frac{1}{8}=1-\frac{1}{2^3} \\
& \therefore t_n=1-\frac{1}{2^n}=1-2^{-n}
\end{aligned}
$
Question 15 .
The sum up to $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+$ is
(a) $\frac{n(n+1)}{2}$
(b) $2 n(n+1)$
(c) $\frac{n(n+1)}{\sqrt{2}}$
(d) 1
Solution:
(c) $\frac{n(n+1)}{\sqrt{2}}$
Hint. $\sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+\ldots n$ terms
$
=\sqrt{2}(1+2+3+\ldots n)=\sqrt{2} \frac{(n)(n+1)}{2}=\frac{n(n+1)}{\sqrt{2}}
$
Question 16.
The value of the series $\frac{1}{2}+\frac{7}{4}+\frac{13}{8}+\frac{19}{16}+\ldots$ is
(a) 14
(b) 7
(c) 4
(d) 6
Solution:

(a) 14
Hint. $\frac{1}{2}+\frac{7}{4}+\frac{13}{8}+\frac{19}{16}+$
It is an arithmetico geometric series
Here $a=1, d=7-1=6$, and $r=\frac{1}{2}$
$
S_{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^2}=\frac{1}{1-1 / 2}+\frac{(6)(1 / 2)}{(1-1 / 2)^2}=\frac{1}{1 / 2}+\frac{3}{(1 / 2)^2}=2+(3 \times 4)=2+12=14
$
Question 17.
The sum of an infinite GP is 18 . If the first term is 6 , the common ratio is .......
(a) $\frac{1}{3}$
(b) $\frac{2}{3}$
(c) $\frac{1}{6}$
(d) $\frac{3}{4}$
Solution:
(b) $\frac{2}{3}$
Hint:
$
\begin{aligned}
& \frac{6}{1-r}=18 \Rightarrow 6=18-18 r \\
& 18 r=18-6=12 \\
& r=12 / 18=2 / 3
\end{aligned}
$
Question 18.
The coefficient of $\mathrm{x}^5$ in the series $\mathrm{e}^{-2 \mathrm{x}}$ is .......
(a) $\frac{2}{3}$
(b) $\frac{3}{2}$
(c) $\frac{-4}{15}$
(d) $\frac{4}{15}$
Answer:
(c) $\frac{-4}{15}$
Hint. $e^{-2 x}=1-\frac{2 x}{\angle 1}+\frac{(2 x)^2}{\angle 2}-\frac{(2 x)^3}{\angle 3}+\frac{(2 x)^4}{\angle 4}-\frac{(2 x)^5}{\angle 5} \cdots$
Coefficient of $x^5=\frac{-2^5}{\angle 5}=\frac{-32}{120}=\frac{-4}{15}$

Question 19.
The value of $\frac{1}{2 !},+\frac{1}{4 !}+\frac{1}{6 !}+\ldots \ldots .$. is
Answer:
(c) $\frac{(e-1)^2}{2 e}$
Hint. $e^1=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !} \cdots, e^{-1}=1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !} \ldots$
$
\begin{aligned}
& \frac{e^1+e^{-1}}{2}=1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots . \\
& \frac{1}{2 !}+\frac{1}{4 !}+\ldots=\frac{e^e+e^{-1}}{2}-1 \\
& =\frac{e^1-2+e^{-1}}{2}=\frac{(e-1)^2}{2}
\end{aligned}
$
Question 20.
The value of $1-\frac{1}{2}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^2-\frac{1}{4}\left(\frac{2}{3}\right)^3+\ldots$ is
(a) $\log \left(\frac{5}{3}\right)$
(b) $\frac{3}{2} \log \left(\frac{5}{3}\right)$
(c) $\frac{5}{3} \log \left(\frac{5}{3}\right)$
(d) $\frac{2}{3} \log \left(\frac{2}{3}\right)$
Solution:
(b) $\frac{3}{2} \log \left(\frac{5}{3}\right)$
$
\text { Hint. } \begin{aligned}
\log (1+x) & =x-\frac{x^2}{2}+\frac{x^3}{3}+\ldots \\
\log \frac{(1+x)}{x} & =1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots \\
\text { put } x=2 / 3 & =1-\frac{(2 / 3)}{2}+\frac{(2 / 3)^2}{3}-\frac{(2 / 3)^3}{4}+\ldots \\
& =\log (1+2 / 3) / 2 / 3 \\
& =\frac{3}{2} \log (1+2 / 3)=\frac{3}{2} \log \frac{5}{3}
\end{aligned}
$