Question

Let $a_1,a_2,a_3,...$ be an $A.P$, such that $\frac{a_1+a_2+\dots +a_p}{a_1+a_2+a_3+\dots +a_q}=\frac{p^3}{q^3};p\ne q$ Then $\frac{a_6}{a_{21}}$ is equal to:

• A. $\frac{41}{11}$
• B. $\frac{31}{121}$
• C. $\frac{11}{41}$
• D. $\frac{121}{1681}$

Answer 1

Answer: (D)

$\frac{a_1+a_2+\dots \dots +a_p}{a_1+a_2+a_3+\dots \dots +a_q}=\frac{p^3}{q^3}$
$\begin{array}{rl}\Rightarrow & \frac{\frac{p}{2}[2a+(p-1)d]}{\frac{q}{2}[2a+(q-1)d]}=\frac{p}{q^3}\\ \Rightarrow & \frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p^2}{q^2}\end{array}$
putting $p=11$ and $q=41$
$\Rightarrow \frac{2a+10d}{2a+40d}={\left(\frac{11}{41}\right)}^2$
$\Rightarrow \frac{a+5d}{a+2d}=\frac{121}{1681}$
$\Rightarrow \frac{a_6}{a_{21}}=\frac{121}{1681}$