Question

Let $a_1,a_2,a_3,\dots$. be the terms of an A.P. . If $\frac{a_1+a_2+\dots +a_p}{a_1+a_2+\dots .+a_q}=\frac{p^2}{q^2},p\ne q$, 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, $\frac{\frac{p}{2}\left[2a_1+(p-1)d\right]}{\frac{q}{2}\left[2a_1+(q-1)d\right]}=\frac{p^2}{q^2}$
$\Rightarrow \frac{\left(2a_1-d\right)+pd}{\left(2a_1-d\right)+qd}=\frac{p}{q}\Rightarrow \left(2a_1-d\right)(p-q)=0$
$\Rightarrow a_1=\frac{d}{2}(\because p\ne q)$
Now $\frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}=\frac{\frac{d}{2}+5d}{\frac{d}{2}+20d}=\frac{11}{41}$