Question

Let $a_1,a_2,a_3,...$ be the terms of an $A.P.$ If $\frac{a_1+a_2+...+a_p}{a_1+a_2+...+a_q}=\frac{p^2}{q^2};p\ne q,$ then $\frac{a_6}{a_{21}}$ is equal to

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

Answer 1

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

Given : $\frac{a_1+a_2+...+a_p}{a_1+a_2+...+a_q}=\frac{p^2}{q^2}$
$\Rightarrow \frac{\frac{p}{2}[2a_1+(p-1\left)d\right]}{\frac{q}{2}[2a_1+(q-1\left)d\right]}=\frac{p^2}{q^2}\Rightarrow \frac{2a_1+(p-1)d}{2a_1+(q-1)d}=\frac{p}{q}\Rightarrow [2a_1+(p-1\left)d\right]q=[2a_1+(q-1\left)d\right]p\Rightarrow 2a_1(q-p)=d\left[\right(q-1)p-(p-1\left)q\right]\Rightarrow 2a_1=d$

$\therefore \frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}=\frac{a_1+10a_1}{a_1+40a_1}=\frac{11}{41}$