Question

The ratio of sum of $m$ and $n$ terms of an $A.P$ is $m^2:n^2$, then the ratio of $m^{th}$ and $n^{th}$ term.

Answer 1

Let the first term be a and common difference be d.

Given: $\frac{S_m}{S_n}=\frac{m^2}{n^2}$

To find: $\frac{T_m}{T_n}$

$\frac{S_m}{S_n}=\frac{\frac{m}{2}[2a+(m-1\left)d\right]}{\frac{n}{2}[2a+(n-1\left)d\right]}=\frac{m^2}{n^2}$

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

Put $m=2m-1$ and $n=2n-1$

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

$\frac{2[a+(m-1\left)\right]}{2[a+(n-1\left)d\right]}=\frac{2m-1}{2n-1}$

$\frac{T_m}{T_n}=\frac{2m-1}{2n-1}$.