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 will be ?

• A. $\frac{m-1}{n-1}$
• B. $\frac{n-1}{m-1}$
• C. $\frac{2m-1}{2n-1}$
• D. $\frac{2n-1}{2m-1}$

Answer 1

Answer: (C)

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

$\frac{S_m}{S_n}=\frac{m^2}{n^2}\frac{\frac{m}{2}[2a+(m-1\left)d\right]}{\frac{n}{2}[2a+(n-1\left)d\right]}=\frac{m^2}{n^2}[2a+(m-1\left)d\right]n=[2a+(n-1\left)d\right]m2a=d⟶\left(1\right)\frac{a_m}{a_n}=\frac{a+(m-1)d}{a+(n-1)d}\left\{\text{as per eq 1}\right\}\frac{a_m}{a_n}=\frac{a+(m-1)2a}{a+(n-1)2a}\frac{a_m}{a_n}=\frac{2m-1}{2n-1}$