Let a_1,a_2,a_3, be an A.P.

Question

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

Accepted Answer

Taking $p=2,q=1$: $\dfrac{a_1+a_2}{a_1}=\dfrac{8}{1}\Rightarrow a_1+(a_1+d)=8a_1\Rightarrow d=6a_1$.

$$\frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}=\frac{a_1+30a_1}{a_1+120a_1}=\frac{31}{121}.$$

Solutions · 1