Question

Let $a_1,a_2,a_3\cdots$ be terms of A.P.
$If\frac{a_1+a_2+\cdots a_p}{a_1+a_2+\cdots +a_q}=\frac{p^2}{q^2},p\ne q,$then $\frac{a_6}{a_{21}}$ equal

• A. $\frac{41}{11}$
• B. $\frac{7}{2}$
• C. $\frac{2}{7}$
• D. $\frac{11}{41}$

Answer 1

The correct option is D $\frac{11}{41}$

$\frac{\frac{p}{2}[2a_1+(p-1\left)d\right]}{\frac{q}{2}[2a_1+(q-1\left)d\right]}=\frac{p^2}{q_2}\Rightarrow \frac{2a_1+(p-1)d}{2a_1+(q-1)d}=\frac{p}{q}\frac{a_1+\left(\frac{p-1}{2}\right)d}{a_1+\left(\frac{q-1}{2}\right)d}=\frac{p}{q}For\frac{a_6}{a_{21}},p=11,q=41\Rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}$