Question

In an A.P. if $\frac{S_m}{S_n}=\frac{m^4}{n^4}$ then prove that $\frac{T_{m+1}}{T_{n+1}}=\frac{(2m+1{)}^3}{(2n+1{)}^3}$.

Answer 1

Given relation implied that
$\frac{\frac{m}{2}[2a+(m-1\left)d\right]}{\frac{n}{2}[2a+(n-1\left)d\right]}=\frac{m^4}{n^4}$
or $\frac{a+\frac{m-1}{2}d}{a+\frac{n-1}{2}d}=\frac{m^3}{n^3}$ $\left(1\right)$
Now $\frac{T_{M+1}}{T_{N+1}}=\frac{a+Md}{a+Nd}$ where
$M=\frac{m-1}{2};N=\frac{n-1}{2}$
or $m=2M+1,n=2N+1$
$\therefore \frac{T_{M+1}}{T_{N+1}}=\frac{(2M+1{)}^3}{(2N+1{)}^3}$ or $\frac{T_{m+1}}{T_{n+1}}=\frac{(2m+1{)}^3}{(2n+1{)}^3}$.