Question

If the ratio of sum of $n$ terms of $2$ different $A.P.$ is $\frac{2n-1}{5n+10}$, then the ratio of their ${15}^{th}$ term is

• A. $\frac{155}{57}$
• B. $\frac{150}{57}$
• C. $\frac{57}{155}$
• D. $\frac{57}{150}$

Answer 1

The correct option is C $\frac{57}{155}$

Given : $\frac{S_n}{S_n^{\prime }}=\frac{2n-1}{5n+10}$
$\Rightarrow \frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2a^{\prime }+(n-1\left)d^{\prime }\right]}=\frac{2n-1}{5n+10}$
$\Rightarrow \frac{a+\frac{(n-1)}{2}d}{a^{\prime }+\frac{(n-1)}{2}{d}^{\prime }}=\frac{2n-1}{5n+10}...\left(1\right)$

We know that
$\frac{T_n}{T_n^{\prime }}=\frac{a+(n-1)d}{a^{\prime }+(n-1)d^{\prime }}$
So, replace $n\to 2n-1$, in $\left(1\right)$ we get
$\frac{T_n}{T_n^{\prime }}=\frac{2(2n-1)-1}{5(2n-1)+10}\Rightarrow \frac{T_n}{T_n^{\prime }}=\frac{4n-3}{10n+5}$

Putting $n=15$, we get
$\frac{T_{15}}{T_{15}^{\prime }}=\frac{57}{155}$


Alternate solution :
Given : $\frac{S_n}{S_n^{\prime }}=\frac{2n-1}{5n+10}$
$\Rightarrow \frac{S_n}{S_n^{\prime }}=\frac{2n^2-n}{5n^2+10n}$
So,
$S_n=(2n^2-n)k,S_n^{\prime }=(5n^2+10n)k\Rightarrow \frac{T_n}{T_n^{\prime }}=\frac{S_n-S_{n-1}}{S_n^{\prime }-S_{n-1}^{\prime }}\Rightarrow \frac{T_n}{T_n^{\prime }}=\frac{\left[2\right\{n^2-(n-1{)}^2\}-\{n-(n-1\left)\right\}]k}{\left[5\right\{n^2-(n-1{)}^2\}+10\{n-(n-1\left)\right\}]k}\Rightarrow \frac{T_n}{T_n^{\prime }}=\frac{2(2n-1)-1}{5(2n-1)+10}\Rightarrow \frac{T_n}{T_n^{\prime }}=\frac{4n-3}{10n+5}$

Putting $n=15$, we get
$\frac{T_{15}}{T_{15}^{\prime }}=\frac{57}{155}$