Question

Let $a_1, a_2, a_3, \ldots$. be the terms of an A.P. . If $\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\ldots .+a_q}=\frac{p^2}{q^2}, p \neq q$, then $\frac{a_6}{a_{21}}$ equals to

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

Answer 1

Answer: (C)

Given, $\frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}=\frac{p^2}{q^2}$
$\Rightarrow \frac{\left(2 a_1-d\right)+p d}{\left(2 a_1-d\right)+q d}=\frac{p}{q} \Rightarrow \left(2 a_1-d\right)(p-q)=0$
$\Rightarrow a_1=\frac{d}{2} (\because p \neq q)$
Now $ \frac{a_6}{a_{21}}=\frac{a_1+5 d}{a_1+20 d}=\frac{\frac{d}{2}+5 d}{\frac{d}{2}+20 d}=\frac{11}{41}$