Exercise 5.2 - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 5.2
Question 1.
Write the first 6 terms of the sequences whose $\mathrm{n}^{\text {th }}$ terms are given below and classify them as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic progression and none of them.
(i) $\frac{1}{2^{n+1}}$
(ii) $\frac{(n+1)(n+2)}{n+3(n+4)}$
(iii) $4\left(\frac{1}{2}\right)^n$
(iv) $\frac{(-1)^n}{n}$
(v) $\frac{2 n+3}{3 n+4}$
(vi) $\frac{3 n-2}{3^{n-1}}$
Solution:
(i) $\begin{aligned} t_n & =\frac{1}{2^{n+1}} \\ t_1 & =\frac{1}{2^{1+1}}=\frac{1}{2^2} ; t_2=\frac{1}{2^3} ; t_3=\frac{1}{2^4} ; t_4=\frac{1}{2^5} ; t_5=\frac{1}{2^6} ; t_6=\frac{1}{2^7}\end{aligned}$
So, the first six terms are $\frac{1}{2^2}, \frac{1}{2^3}, \frac{1}{2^4}, \frac{1}{2^5}, \frac{1}{2^6}$ and $\frac{1}{2^7}$ which is a G.P. with $a=\frac{1}{2^2}$ and $r=\frac{1}{2}$.

(ii)
$
\begin{aligned}
& t_n=\frac{(n+1)(n+2)}{n+3(n+4)} \\
& t_1=\frac{(1+1)(1+2)}{1+3(1+4)}=\frac{2 \times 3}{1+15}=\frac{6}{16}=\frac{3}{8} \\
& t_2=\frac{(3)(4)}{2+3(6)}=\frac{12}{2+18}=\frac{12}{20}=\frac{3}{5} \\
& t_3=\frac{(4)(5)}{3+3(7)}=\frac{20}{24}=\frac{5}{6} \\
& t_4=\frac{(5)(6)}{4+3(8)}=\frac{30}{4+24}=\frac{30}{28}=\frac{15}{14} \\
& t_5=\frac{(6)(7)}{5+3(9)}=\frac{42}{32}=\frac{21}{16} \\
& t_6=\frac{(7)(8)}{6+3(10)}=\frac{56}{36}=\frac{14}{9}
\end{aligned}
$

So the first six terms are $\frac{3}{8}, \frac{3}{5}, \frac{5}{6}, \frac{15}{14}, \frac{21}{16}, \frac{14}{9}$.
It is not a G.P. or A.P. or H.P. or A.G.P.

(iii)
$
\begin{aligned}
& t_n=4\left(\frac{1}{2}\right)^n \\
& t_1=4\left(\frac{1}{2}\right)^1=2 \\
& t_2=4\left(\frac{1}{2}\right)^2=4 \times \frac{1}{4}=1 \\
& t_3=4\left(\frac{1}{2}\right)^3=\frac{4}{8}=\frac{1}{2} ; t_4=4\left(\frac{1}{2}\right)^4=\frac{4}{16}=\frac{1}{4} \\
& t_5=4\left(\frac{1}{2}\right)^5=\frac{4}{32}=\frac{1}{8} ; t_6=4\left(\frac{1}{2}\right)^6=\frac{4}{64}=\frac{1}{16}
\end{aligned}
$
So the first six terms are $2,1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}$ and $\frac{1}{16}$ which is a G.P. with $a=2$ and $r=\frac{1}{2}$.

(iv)

$
\begin{aligned}
& t_n=\frac{(-1)^n}{n} \\
& t_1=\frac{(-1)^1}{1}=-1 \quad ; \quad t_2=\frac{(-1)^2}{2}=\frac{1}{2} \\
& t_3=\frac{(-1)^3}{3}=-\frac{1}{3} \quad ; \quad t_4=\frac{(-1)^4}{4}=\frac{1}{4} \\
& t_5=\frac{(-1)^5}{5}=-\frac{1}{5} \quad ; \quad t_6=\frac{(-1)^6}{6}=\frac{1}{6}
\end{aligned}
$
So the first 6 terms are $-1, \frac{1}{2},-\frac{1}{3}, \frac{1}{4},-\frac{1}{5}, \frac{1}{6}$.
It is not an A.P. or G.P. or H.P. or A.G.P

(v) $t_n=\frac{2 n+3}{3 n+4}$
$
\begin{aligned}
& t_1=\frac{2+3}{3+4}=\frac{5}{7} \quad ; \quad t_2=\frac{4+3}{6+4}=\frac{7}{10} \\
& t_3=\frac{6+3}{9+4}=\frac{9}{13} \quad ; \quad t_4=\frac{8+3}{12+4}=\frac{11}{16} \\
& t_5=\frac{10+3}{15+4}=\frac{13}{19} \quad ; \quad t_6=\frac{12+3}{18+4}=\frac{15}{22} \\
&
\end{aligned}
$
So the first 6 terms are $\frac{5}{7}, \frac{7}{10}, \frac{9}{13}, \frac{11}{16}, \frac{13}{19}$ and $\frac{15}{22}$.
It is not an A.P. or G.P. or H.P. or A.G.P.

(vi) $t_n=\frac{3 n-2}{3^{n-1}}$
$
\begin{aligned}
t_1 & =\frac{3-2}{3^0}=1 & ; & t_2=\frac{6-2}{3^1}=\frac{4}{3} \\
t_3 & =\frac{9-2}{3^2}=\frac{7}{9} & ; & t_4=\frac{12-2}{3^3}=\frac{10}{27} \\
t_5 & =\frac{15-2}{3^4}=\frac{13}{81} & ; & t_6=\frac{18-2}{3^5}=\frac{16}{243}
\end{aligned}
$
So the first 6 terms are $1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \frac{13}{81}$ and $\frac{16}{243}$ i.e., $\frac{1}{3^0}, \frac{4}{3^1}, \frac{7}{3^2}, \frac{10}{3^3}, \frac{13}{3^4}, \frac{16}{3^5}$.
It is a A.G.P.

Question 2.
Write the first 6 terms of the sequences whose $n^{\text {th }}$ term $a_n$ is given below.
(i) $a_n= \begin{cases}n+1 & \text { if } n \text { is odd } \\ n & \text { if } n \text { is even }\end{cases}$
Solution:
$
\begin{aligned}
& a_1=1+1=2 ; a_2=2 \\
& a_3=3+1=4 ; a_4=4 \\
& a_5=5+1=6 ; a_6=6
\end{aligned}
$
So, the first 6 terms are $2,2,4,4,6,6$
(ii) $a_n=\left\{\begin{array}{lr}1 & \text { if } n=1 \\ 2 & \text { if } n=2 \\ a_{n-1}+a_{n-2} & \text { if } n>2\end{array}\right.$
Solution:
$
\begin{aligned}
& a_1=1 ; a_2=2, a_3=3 \\
& a_4=a_3+a_2+a_1=3+2+1=6 \Rightarrow a_4=6 \\
& a_5=a_4+a_3+a_2=6+3+2=11 \Rightarrow a_5=11 \\
& a_6=a_5+a_4+a_3=11+6+3=20 \Rightarrow a_6=20
\end{aligned}
$
So the first 6 terms are $1,2,3,5,8,13$.
(iii) $a_n= \begin{cases}n & \text { if } n \text { is } 1 \\ a_{n-1}+a_{n-2}+a_{n-3} & \text { if } n>3\end{cases}$

Solution:
$
\begin{aligned}
& a_1=1 \quad ; a_2=2 \quad ; \quad a_3=3 \\
& a_4=a_3+a_2+a_1=3+2+1=6 \quad \Rightarrow \quad \Rightarrow \quad a_4=6 \\
& a_5=a_4+a_3+a_2=6+3+2=11 \quad \Rightarrow a_5=11 \\
& a_6=a_5+a_4+a_3=11+6+3=20 \Rightarrow a_6=20
\end{aligned}
$
So the first 6 terms are $1,2,3,6,11,20$.
Question 3.
Write the $n_{\text {th }}$ term of the following sequences.
Solution:
(i) $2,2,4,4,6,6 \ldots \ldots$
Solution:
$a_n= \begin{cases}n+1 & \text { if } n \text { is odd } \\ n & \text { if } n \text { is even }\end{cases}$
(ii) $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots$
Solution:
Nr: $1,2,3, \ldots \ldots \mathrm{t}_{\mathrm{n}}=\mathrm{n}$
Dr: $2,3,4, \ldots . \mathrm{t}_{\mathrm{n}}=\mathrm{n}+1$
So the $n^{\text {th }}$ term is $t_n=\frac{n}{n+1} \forall n \in \mathrm{N}$
(iii) $\frac{1}{2}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \frac{9}{10}, \ldots$
Solution:
Nr: $1,3,5,7, \ldots$ which is an A.P. $a=1, d=3-1=2$
$\mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$\mathrm{t}_{\mathrm{n}}=1+(\mathrm{n}-1) 2=1+2 \mathrm{n}-2=2 \mathrm{n}-1$.
Dr : $2,4,6,8, \ldots$
So the $\mathrm{n}^{\text {th }}$ term is $2+(\mathrm{n}-1) 2=2+2 \mathrm{n}-2=2 \mathrm{n}$.
$
\therefore t_n=\frac{2 n-1}{2 n}, \forall n \in \mathrm{N}
$

(iv) $6,10,4,12,2,14,0,16,-2, \ldots$.
Solution:
$
\begin{aligned}
& t_1=6 ; t_2=10 \\
& t_3=4 ; t_4=12 \\
& t_5=2 ; t_6=14 \\
& t_7=0 ; t_8=16
\end{aligned}
$
When $\mathrm{n}$ is odd, the sequence is $6,4,2,0, \ldots$
(i.e.) $a=6$ and $d=4-6=-2$.
So, $\mathrm{t}_{\mathrm{n}}=6+(\mathrm{n}-1)(-2)=6-2 \mathrm{n}+2=8-2 \mathrm{n}$
When $\mathrm{n}$ is even, the sequence is $10,12,14,16, \ldots$
Here $\mathrm{a}=10$ and $\mathrm{d}=12-10=2$
$
t_n=10+(n-1) 2=10+2 n-2=2 n+8 \text { (i.e.) } 8+2 n
$
$\therefore t_n=\left\{\begin{array}{l}8-2 n \text { when } n \text { is odd } \\ 8+2 n \text { when } n \text { is even }\end{array} \quad\right.$ or $t_n= \begin{cases}7-n & \text { when } n \text { is odd } \\ 8+n & \text { when } n \text { is even }\end{cases}$

Question 4.
The product of three increasing numbers in GP is 5832 . If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP.
Solution:
The 3 numbers in a G.P. is taken as $\frac{a}{r}$, a, ar
Their product is 5832 .
$
\Rightarrow \frac{a}{r} \times a \times a r=5832 \text { (i.e.) } a^3=5832=18^3 \Rightarrow a=18
$
$\therefore$ The 3 numbers are $\frac{18}{r}, 18,18 r$
When 6 added to $t_2$ and 9 added to $t_3$ we get $\frac{18}{r}, 24,18 r+9$ which is an A.P.
$
\begin{aligned}
& \quad \frac{18}{r}, 24,18 r+9 \text { are in A.P. } \\
& \Rightarrow \quad 24-\frac{18}{r}=18 r+9-24 \\
& \text { (i.e.) } \quad 24-9+24=18 r+\frac{18}{r} \\
& \quad 18 r+\frac{18}{r}=39 \\
& (\div \text { by } 3) 6 r+\frac{6}{r}=13 \\
& 6 \mathrm{r}^2+6=13 \\
& 6 \mathrm{r}^2-13 \mathrm{r}+6=0 \\
& (3 \mathrm{r}-2)(2 \mathrm{r}-3)=0 \\
& \mathrm{r}=2 / 3 \text { or } 3 / 2
\end{aligned}
$
When $a=18$ and $r=2 / 3$, the G.P. is $18,18 \times \frac{2}{3}, 18 \times\left(\frac{2}{3}\right)^2, \ldots \ldots$.
(i.e.) $18,12,8, \ldots$
When $a=18$ and $r=3 / 2$, the G.P. is $18,18 \times \frac{3}{2}, 18 \times \frac{3}{2} \times \frac{3}{2}, \ldots \ldots$
(i.e.) $18,2 \dot{7}, \frac{81}{2}, \ldots$
Question 5.
Write the $n^{\text {th }}$ term of the sequence $\frac{3}{1^2 2^2}, \frac{5}{2^2 3^2}, \frac{7}{3^2 4^2}, \ldots$ as a difference of two terms.

Solution:
$
\begin{aligned}
& t_1=\frac{3}{1^2 2^2}, t_2=\frac{5}{2^2 3^2}, t_3=\frac{7}{3^2 4^2} \\
& \mathrm{Nr}: 3,5,7, \ldots\{\text { A.P. } a=3, d=5-3=2\} \\
& t_n=3+(n-1) 2=3+2 n-2=2 n+1 \\
& \text { Dr : } 1^2 2^2, 2^2 3^2, \ldots \\
& \text { So } t_n=n^2(n+1)^2 \\
& \therefore n^{\text {th }} \text { term }=\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}
\end{aligned}
$
Question 6.
If $\mathrm{t}_{\mathrm{k}}$ is the $\mathrm{k}^{\text {th }}$ term of a G.P., then show that $\mathrm{t}_{\mathrm{n}-\mathrm{k}}, \mathrm{t}_{\mathrm{n}}, \mathrm{t}_{\mathrm{n}}+\mathrm{k}$ also form a GP for any positive integer $\mathrm{k}$. Solution:
Let a be the first term and $\mathrm{r}$ be the common ratio.
We are given $t_k=a r^{k-1}$
We have to Prove : $t_{n-k}, t_n, t_n+k$ form a G.P.
$
\begin{aligned}
t_{n-k} & =a r^{n-k-1} \\
t_n & =a r^{n-1} \\
t_{n+k} & =a r^{n+k-1}
\end{aligned}
$
Now $\quad \frac{t_n}{t_{n-k}}=\frac{a r^{n-1}}{a r^{n-k-1}}=r^{n-1-n+k+1}=r^k$
Also $\quad \frac{t_{n+k}}{t_n}=\frac{a r^{n+k-1}}{a r^{n-1}}=r r^{n+k-1-n+1}=r^k$
Now $\quad \frac{t_n}{t_{n-k}}=\frac{t_{n+k}}{t_n}$
$\Rightarrow \quad t_{n-k}, t_n, t_{n+k}$ form a G.P.

Question 7.
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in geometric progression, and if $a^{\frac{1}{2}}=b^{\frac{1}{y}}=c^{\frac{1}{2}}$, then prove that $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in arithmetic progression.
Solution:
Given a, b, c are in G.P.
$
\begin{aligned}
& \Rightarrow b^2=a c \\
& \Rightarrow \log b^2=\log a c \\
& \text { (i.e.) } 2 \log b=\log a+\log c .
\end{aligned}
$
We are given $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k$ (say)
$
\begin{aligned}
& \Rightarrow \quad \log a^k=\frac{1}{x} ; \log b^k=\frac{1}{y} ; \log c^k=\frac{1}{7} \\
& \Rightarrow \quad \log a^k=\frac{1}{x} \Rightarrow x=\log k^a \\
& \text { Similarly } \quad y=\log k^b \\
& z=\log k^c \\
&
\end{aligned}
$
Substituting these values in equation (1) we get $2 y=x+z \Rightarrow x, y z$ are in A.P.

Question 8.
The AM of two numbers exceeds their GM by 10 and $\mathrm{HM}$ by 16 . Find the numbers.
Solution:
Let the two numbers be $a$ and $b$.
Their A.M. $=\frac{a+b}{2}$
G.M. $=\sqrt{a b}$ and H.M. $=\frac{2 a b}{a+b}$
We are given A.M. = G.M. $+10=$ H.M. +16
(i.e.) $\frac{a+b}{2}=\sqrt{a b}+10$
and $\frac{a+b}{2}=\frac{2 a b}{a+b}+16$
from (2) $\frac{a+b}{2}-\frac{2 a b}{a+b}=16$
$
\Rightarrow(a+b)^2-4 a b=16(2)(a+b)
$
(i.e.) $\quad(a-b)^2=32(a+b)$
$
\begin{aligned}
& (1) \Rightarrow \frac{a+b}{2}=\sqrt{a b}+10 \\
& \Rightarrow \quad a+b=2 \sqrt{a b}+20 \\
& \Rightarrow \quad a+b-20=2 \sqrt{a b}
\end{aligned}
$
So, $(a+b-20)^2=4 a b$
$
\begin{aligned}
& \text { (i.e.) }(a+b)^2+400-40(a+b)=4 a b \\
& (a+b)^2-4 a b=40(a+b)-400 \\
& \text { from(3) }(a+b)^2-4 a b=32(a+b) \\
& \Rightarrow 32(a+b)=40(a+b)-400 \\
& (-b y 8) 4(a+b)=5(a+b)-50 \\
& 4 a+4 b=5 a+5 b-50 \\
& a+b=50 \\
& a=50-b
\end{aligned}
$

Substituting $a=50-\mathrm{b}$ in (3) we get
$
\begin{aligned}
& (50-\mathrm{b}-\mathrm{b})^2=32(50) \\
& (50-2 b)^2=32 \times 50 \\
& {[2(25-b)]^2=32 \times 50} \\
& \text { (i.e.) } \quad 4(25-b)^2=32 \times 50 \\
& \Rightarrow \quad(25-b)^2=\frac{32 \times 50}{4}=8 \times 50=400=20^2 \\
& \Rightarrow \quad 25-b= \pm 20 \\
&
\end{aligned}
$

When $b=5, a=50-5=45$
When $b=45, \mathrm{a}=50-45=5$
So the two numbers are 5 and 45 .
Question 9.
If the roots of the equation $(q-r) x^2+(r-p) x+p-q=0$ are equal, then show that $p, q$ and $r$ are in $A P$.

Solution:
The roots are equal $\Rightarrow \Delta=0$
(i.e.) $b^2-4 a c=0$
Hence, $a=q-r ; b=r-p ; c=p-q$
$\mathrm{b}^2-4 \mathrm{ac}=0$
$\Rightarrow(\mathrm{r}-\mathrm{p})^2-4(\mathrm{q}-\mathrm{r})(\mathrm{p}-\mathrm{q})=0$
$\mathrm{r}^2+\mathrm{p}^2-2 \mathrm{pr}-4\left[\mathrm{qr}-\mathrm{q}^2-\mathrm{pr}+\mathrm{pq}\right]=0$
$\mathrm{r}^2+\mathrm{p}^2-2 \mathrm{pr}-4 \mathrm{qr}+4 \mathrm{q}^2+4 \mathrm{pr}-4 \mathrm{pq}=0$
(i.e.) $p^2+4 q^2+r^2-4 p q-4 q r+2 p r=0$
(i.e.) $(p-2 q+r)^2=0$
$\Rightarrow \mathrm{p}-2 \mathrm{q}+\mathrm{r}=0$
$\Rightarrow \mathrm{p}+\mathrm{r}=2 \mathrm{q}$
$\Rightarrow \mathrm{p}, \mathrm{q}, \mathrm{r}$ are in A.P.
Question 10.
If $a, b, c$ are respectively the $\mathrm{p}^{\text {th }}, \mathrm{q}^{\text {th }}$ and $\mathrm{r}^{\text {th }}$ terms of a G.P., show that $(\mathrm{q}-\mathrm{r}) \log a+(r-p) \log \mathrm{b}+(\mathrm{p}-$ q) $\log c=0$.
Solution:
Let the G.P. be $1, \mathrm{lk}, 1 \mathrm{lk}^2, \ldots$
We are given $t_p=a, t_q=b, t_{\mathrm{r}}=c$

$\begin{aligned}
& \Rightarrow a=l k^{p-1} ; b=l k^{q-1} ; c=l k^{r-1} \\
& a=l k^{p-1} \quad \Rightarrow \quad \log a=\log l+\log k^{p-1}=\log l+(p-1) \log k \\
& b=l k^{q-1} \quad \Rightarrow \quad \log b=\log l+\log k^{q-1}=\log l+(q-1) \log k \\
& c=l k^{r-1} \quad \Rightarrow \quad \log c=\log l+\log k^{r-1}=\log l+(r-1) \log k \\
& \text { LHS }=(\mathrm{q}-\mathrm{r}) \log \mathrm{a}+(\mathrm{r}-\mathrm{p}) \log \mathrm{b}+(\mathrm{p}-\mathrm{q}) \log \mathrm{c} \\
& =(q-r)[\log l+(p-1) \log k]+(r-p)[\log l+(q-1) \log k]+ \\
& (p-q)[\log l+(r-1) \log k] \\
& =\log l[p-q+q-r+r-p]+\log k[(q-r)(p-1)+(r-p)(q-1)+ \\
& (p-q)(r-1)] \\
& =\log l(0)+\log k[p(q-r)+q(r-p)+r(p-q)-(q-r+r-p+p-q)] \\
& =0=\text { RHS } \text {. } \\
&
\end{aligned}$