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Let $a_1, a_2, a_3, \ldots$ be terms of an A.P. If $\frac{a_1+a_2+\cdots a_p}{a_1+a_2+\cdots+a_q}=\frac{p^2}{q^2}, p \neq q$, then $\frac{a_6}{a_{21}}$ equals
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Verified Answer
The correct answer is:
$\frac{11}{41}$
$\frac{11}{41}$
$$
\begin{aligned}
& \frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}=\frac{p^2}{q^2} \Rightarrow \frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}=\frac{p}{q} \\
& \frac{a_1+\left(\frac{p-1}{2}\right) d}{a_1+\left(\frac{q-1}{2}\right) d}=\frac{p}{q}
\end{aligned}
$$
For $\frac{a_6}{a_{21}}, p=11, q=41 \rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}$
\begin{aligned}
& \frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}=\frac{p^2}{q^2} \Rightarrow \frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}=\frac{p}{q} \\
& \frac{a_1+\left(\frac{p-1}{2}\right) d}{a_1+\left(\frac{q-1}{2}\right) d}=\frac{p}{q}
\end{aligned}
$$
For $\frac{a_6}{a_{21}}, p=11, q=41 \rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}$
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