Question

Let $a_1,a_2,a_3$ , ..... be terms of an A.P. if the ratio of sum of first $m$ terms to that of sum of first $n$ terms is equal to $\frac{m^2}{n^2}(m\ne n)$ , then $\frac{a_6}{a_{21}}$ equals

• A. $\frac{11}{41}$
• B. $\frac{2}{7}$
• C. $\frac{41}{11}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A $\frac{11}{41}$

sum of first m terms =$\frac{m}{2}2a_1+(m-1)d$

sum of first n terms= $\frac{n}{2}2a_1(n-1)d$ where dis the common differences.

Given,$\frac{\frac{m}{2}2a_1+(m-1)d}{\frac{m}{2}2a_1+(n-1)d}=\frac{m^2}{n^2}$

$\Rightarrow \frac{2a_{1+n(m-1)d}}{2a_1(n-1)d}=\frac{m}{n}$

$\Rightarrow 2n{a}_1+n(m-1)d=2a_{1}m+nmd-md$

$2a_1(n-m)d=2a_1+m(n-1)d$

$\Rightarrow 2a_{1}n+nmd-nd=2a_{1}m+nmd-md$

$\Rightarrow 2a_1(n-m)=(n-m)d$

$\Rightarrow 2a_1=d-----\left(1\right)$

$Now,a_6=a_1+(6-1)d=a_1+5d=a_1+5\left(2a_1\right)$

$=a_1+10a_1$

$=11a_1$

$a_{21}=a_1+(21-1)d=a_1+20d=a_1+20\left(2a_1\right)$

$=a_1+40a_1$

$=41a_1$

$\therefore \frac{a_6}{a_{21}}=\frac{11a_1}{41a_1}=\frac{11}{41}$