# Solve the following

Question:

If $S_{n}$ denotes the sum of first $n$ terms of an A.P.  such that $\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}$, then $\frac{a_{m}}{a_{n}}=$

(a) $\frac{2 m+1}{2 n+1}$

(b) $\frac{2 m-1}{2 n-1}$

(c) $\frac{m-1}{n-1}$

(d) $\frac{m+1}{n+1}$

Solution:

(d) $\frac{m+1}{n+1}$

$\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}$

$\Rightarrow \frac{\frac{m}{2}\{2 a+(m-1) d\}}{\frac{n}{2}\{2 a+(n-1) d\}}=\frac{m^{2}}{n^{2}}$

$\Rightarrow \frac{\{2 a+(m-1) d\}}{\{2 a+(n-1) d\}}=\frac{m}{n}$

$\Rightarrow 2 a n+n d m-n d=2 a m+n m d-m d$

$\Rightarrow 2 a n-2 a m-n d+m d=0$

$\Rightarrow 2 a(n-m)-d(n-m)=0$

$\Rightarrow 2 a(n-m)=d(n-m)$

$\Rightarrow d=2 a \quad \ldots .(1)$

Ratio of $\frac{a_{m}}{a_{n}}=\frac{a+(m-1) d}{a+(n-1) d}$

$\Rightarrow \frac{a_{m}}{a_{n}}=\frac{a+(m-) 2 a}{a+(n-) 2 a} \quad$ From $(1)$

$=\frac{a+2 a m-2 a}{a+2 a n-2 a}$

$=\frac{2 a m-a}{2 a n-a}$

$=\frac{2 m-1}{2 n-1}$