Exercise 5.3-Additional Questions - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
If the ratio of the sums of $m$ terms and $n$ terms of an A.P. be $m^2: n^2$, prove that the ratio of its $m^{\text {th }}$ and $\mathrm{n}^{\text {th }}$ terms is $(2 \mathrm{~m}-1):(2 \mathrm{n}-1)$.
Solution:
Let ' $a$ ' be the first term and $d$ be the common difference of A.P.
Using the given information, we have $\frac{S_m}{S_n}=\frac{m^2}{n^2}$
$
\begin{aligned}
& \frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^2}{n^2} \Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n} \\
\Rightarrow & 2 \mathrm{an}+(\mathrm{mn}-\mathrm{n}) \mathrm{d}=2 \mathrm{am}+(\mathrm{mn}-\mathrm{m}) \mathrm{d} \\
\Rightarrow & 2 \mathrm{an}-2 \mathrm{am}=(\mathrm{mn}-\mathrm{m}-\mathrm{mn}+\mathrm{n}) \mathrm{d} \\
\Rightarrow & 2 \mathrm{a}(\mathrm{n}-\mathrm{m})=(\mathrm{n}-\mathrm{m}) \mathrm{d} \Rightarrow \mathrm{d}=2 \mathrm{a},[\mathrm{n}-\mathrm{m} \neq 0, \mathrm{as} \mathrm{n} \neq \mathrm{m}] \\
& \text { Now, } \frac{a_m}{a_n}=\frac{a+(m-1) d}{a+(n-1) d}=\frac{a+(m-1) \cdot 2 a}{a+(n-1) \cdot 2 a}=\frac{a(1+2 m-2)}{a(1+2 n-2)}=\frac{2 m-1}{2 n-1}
\end{aligned}
$
Hence, the required ratio is $(2 m-1):(2 n-1)$
Question 2.
Determine the number of terms of geometric progression $\left\{a_n\right\}$ if $a_1=3, a_n=96, S_n=189$.
Solution:

Let $r$ be the common ratio of G.P. $a_n=a_1 r^{n-1}=3 r^{n-1}$
But, $a_n=96[$ Given $]$
$
\begin{aligned}
& \therefore 3 r^{n-1}=96 \Rightarrow r^{n-1}=96 \div 3=32 \\
& \therefore r^n=32 r \ldots \ldots \ldots \ldots . .(1)
\end{aligned}
$
Now $S_n=\frac{a_1\left(r^n-1\right)}{r-1} \Rightarrow 189=\frac{3(32 r-1)}{r-1}$
$
\begin{aligned}
& \therefore \frac{32 r-1}{r-1}=189 / 3=63 \\
& \therefore 32 r-1=63(r-1) \Rightarrow 32 r-1=63 r-63 \\
& \therefore 63 r-32 r=63-1 \\
& \Rightarrow 31 r=62 \Rightarrow r=62 / 31=2 \\
& \text { Now, } r^{n-1}=32 \Rightarrow 2^{n-1}=2^5 \\
& \Rightarrow n-1=5, \text { or } n=5+1=6(\text { i.e }) \text { Number of terms }=6
\end{aligned}
$
Question 3.
The sum of first three terms of a G.P. is to the sum of the first six terms as $125: 152$. Find the common ratio of the G.P.
Solution:

Here, $\frac{S_3}{S_6}=\frac{125}{152}$
$
\begin{aligned}
& \Rightarrow \frac{a\left(r^3-1\right) /(r-1)}{a\left(r^6-1\right) /(r-1)}=\frac{125}{152} \Rightarrow \frac{r^3-1}{r^6-1}=\frac{125}{152} \\
& \therefore \frac{r^3-1}{\left(r^3-1\right)\left(r^3+1\right)}=\frac{125}{152} \Rightarrow \frac{1}{r^3+1}=\frac{125}{152} \\
& \therefore 152=125 r^3+125 \Rightarrow 125 r^3=27 \\
& \therefore r^3=\frac{27}{125}=\left(\frac{3}{5}\right)^3 \\
& \Rightarrow r=\left\{\left(\frac{3}{5}\right)^3\right\}^{1 / 3}=\frac{3}{5}
\end{aligned}
$
Hence, the common ratio of the G.P. is $\frac{3}{5}$
Question 4.
Find the sum to $n$ terms the series: $(x+y)+\left(x^2+x y+y^2\right)+\left(x^3+x^2 y+x y^2+y^3\right)+\ldots$
Solution:
$
\begin{aligned}
& (\mathrm{x}+\mathrm{y})+\left(\mathrm{x}^2+\mathrm{xy}+\mathrm{y}^2\right)+\left(\mathrm{x}^3+\mathrm{x}^2 \mathrm{y}+\mathrm{xy}^2+\mathrm{y}^3\right)+\ldots \text { to } \mathrm{n} \text { terms } \\
& =\frac{x^2-y^2}{x-y}+\frac{x^3-y^3}{x-y}+\frac{x^4-y^4}{x-y}+\ldots \text { to } n \text { terms } \\
& =\frac{1}{x-y}\left[\left(x^2-y^2\right)+\left(x^3-y^3\right)+\left(x^4-y^4\right)+\ldots \text { to } n \text { terms }\right] \\
& =\frac{1}{x-y}\left[\left(x^2+x^3+x^4+\ldots \text { to } n \text { terms }\right)-\left[\left(y^2+y^3+y^4+\ldots \text { to } n \text { terms }\right)\right]\right. \\
& =\frac{1}{x-y}\left[\frac{x^2\left(x^n-1\right)}{x-1}-\frac{y^2\left(y^n-1\right)}{y-1}\right]
\end{aligned}
$

Question 5 .
Sum the series: $(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots$ up to $n$ terms
Solution:
$(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots$ up to $n$ terms $=\frac{1-x^2}{1-x}+\frac{1-x^3}{1-x}+\ldots \frac{1-x^4}{1-x}+\ldots$ to $n$ terms
$=\frac{1}{1-x}\left[(1+1+1+\ldots\right.$ to $n$ terms $)-\left(x^2+x^3+x^4 \ldots\right.$ to $n$ terms $)$
$
=\frac{1}{1-x}\left[n-\frac{x^2\left(1-x^n\right)}{1-x}\right]
$

Question 6.
Sum up to $\mathrm{n}$ terms the series: $7+77+777+7777+\ldots$
Solution:
$\mathrm{S}=1+77+777+7777+\ldots$ to $\mathrm{n}$ terms
$
\begin{aligned}
& =\frac{7}{9}[9+99+999+9999+\ldots \text { to } n \text { terms }] \\
& =\frac{7}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\left(10^4-1\right)+\ldots \text { to } n \text { terms }\right] \\
& =\frac{7}{9}\left[\left(10+10^2+10^3+\ldots \text { to } 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}{9}\left(10^n-1\right)-n\right] \\
& =\frac{7}{81}\left[10^{n+1}-9 n-10\right]
\end{aligned}
$
Question 7.
If $S_1, S_2, S_3$ be respectively the sums of $n, 2 n, 3 n$, terms of a G.P. , then prove that $S_1\left(S_3-S_2\right)=\left(S_2-\right.$ $\left.\mathrm{S}_1\right)_2$
Solution:
Let a be the first term and $\mathrm{r}$ be the common ratio of G.P.

$
\therefore S_1=\frac{a\left(r^n-1\right)}{r-1}, S_2=\frac{a\left(r^{2 n}-1\right)}{r-1}, S_3=\frac{a\left(r^{3 n}-1\right)}{r-1}
$
where $r \neq 1$
$
\begin{aligned}
& S_3-S_2=\frac{a}{r-1}\left(r^{3 n}-r^{2 n}\right)=\frac{a\left(r^n-1\right)}{r-1} r^{2 n} \\
& S_1\left(S_3-S_2\right)=\frac{a\left(r^n-1\right)}{r-1} \times \frac{a\left(r^n-1\right)}{7^{2 r-1}} r^{2 n} \\
& =\left[\frac{a\left(\stackrel{r}{r}^n-1\right)}{r-1} \cdot r^n\right] \\
& \left(S_2-S_1\right)=\frac{a}{r-1}\left(r^{2 n}-r^n\right)=\frac{a\left(r^n-1\right)}{r-1} r^n \\
& \therefore S_1\left(S_3-S_2\right)=\left(S_2-S_1\right)^2 \text { [From (1) and (2)] } \\
&
\end{aligned}
$
When $r=1, S_1=n a, S_2=2 n a$ and $S_3=3 n a$
Then, $\left(S_2-S_1\right)^2=(2 n a-n a)^2=n^2 a^2$ and
$
\begin{aligned}
& S_1\left(S_3-S_2\right)=n a(3 n a-2 n a)=n a(n a)=n^2 a^2 \\
& \therefore S_1\left(S_3-S_2\right)=\left(S_2-S_1\right)^2
\end{aligned}
$
Question 8.
If sum of the $n$ terms of a $G . P$ be $S$, their product $P$ and the sum of their reciprocals $R$, the prove that $P^2=\left(\frac{S}{R}\right)^n$
Solution:

Let a be the first term and $\mathrm{r}$ be the common ratio of the G.P.
$
\begin{aligned}
\therefore S & =a+a r+a r^2+\ldots+a r^{n-1} \\
& =\frac{a\left(1-r^n\right)}{1-r} \\
P & =a \times a r \times a r^2 \times \ldots \times a r^{n-1}=a^n r^{1+2+3 \ldots+(n-1)}=a^n r^{n(n-1) / 2} \\
\therefore P^2 & =a^{2 n} r^{n(n-1)} \\
R & =\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}+\ldots+\frac{1}{a r^{n-1}} \\
\Rightarrow R & =\frac{1}{a} \cdot \frac{\left(1-\frac{1}{r^n}\right)}{\left(1-\frac{1}{r}\right)}=\frac{\left(r^n-1\right)}{(r-1)} \cdot \frac{1}{a r^{n-1}}[\because \text { Here, } r<1] \\
\therefore \frac{S}{R} & =a \frac{\left(1-r^n\right)}{1-r} \cdot \frac{r-1}{r^n-1} a r^{n-1}=a^2 r^{(n-1)}
\end{aligned}
$