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

Question

Let $a_1,a_2,a_3,\ldots$ be terms of an A.P. If $\dfrac{a_1+a_2+\cdots+a_p}{a_1+a_2+\cdots+a_q}=\dfrac pq,\ p
e q$, then $\dfrac{a_6}{a_{21}}$ equals:

Accepted Answer

Using $S_{2n-1}=(2n-1)a_n$, take $p=11=2(6)-1$ and $q=41=2(21)-1$:

$$\frac{a_6}{a_{21}}=\frac{S_{11}}{S_{41}}=\frac{p}{q}=\frac{11}{41}.$$

Solutions · 1