Question

The ratio of the 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. $\frac{\left(m-1\right)}{\left(n-1\right)}$
• B. $\frac{(n-1)}{(m-1)}$
• C. $\frac{(2m-1)}{(2n-1)}$
• D. $\frac{(2n-1)}{(2m-1)}$

Answer 1

The correct option is C

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

Explanation of the correct option:

Step 1: Compute the value of $d$.

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

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

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

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

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

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

$\Rightarrow$ $2an+mnd-nd=2am+mnd-md$

$\Rightarrow$ $2an-2am=nd-md$

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

$\Rightarrow$ $2a=d$

Step 2: Compute the required ratio:

Since $k^{th}$ term of A.P. is given by,

$a_k=a+(k-1)d$

Therefore,

$a_m=a+(m-1)d$

$\Rightarrow$$a_m=a+(m-1)2a$

$\Rightarrow$$a_m=a(2m-1)$

$a_n=a+(n-1)d$

$\Rightarrow$$a_n=a+(n-1)2a$

$\Rightarrow$$a_n=a(2n-1)$

Hence, $\frac{a_m}{a_n}=\frac{(2m-1)}{(2n-1)}$

Hence, option (C) is the correct option.