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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \frac{a_1+a_2+a_3+\ldots \ldots+a_p}{a_1+a_2+a_3+\ldots \ldots+a_q}=\frac{p^3}{q^3} \ \Rightarrow & \frac{a_1+a_2}{a_1}=\frac{8}{1} \Rightarrow a_1+\left(a_1+dight)=8 a_1 \ \Rightarrow \quad & d=6 a_1 \ & 	ext { Now } \frac{a_6}{a_{21}}=\frac{a_1+5 d}{a_1+20 d} \ = & \frac{a_1+5 	imes 6 a_1}{a_1+20 	imes 6 a_1}=\frac{1+30}{1+120}=\frac{31}{121}\end{aligned}$

Let $a_1, a_2, a_3, \ldots$ be an A.P, such that…

Solution