Question

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

• A. $\frac{41}{11}$
• B. $\frac{31}{11}$
• C. $\frac{11}{41}$
• D. $\frac{11}{31}$

Answer 1

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

Let $d$ be the common difference of A.P.
$\frac{a_1+a_2+\cdots +a_p}{a_1+a_2+\cdots +a_q}=\frac{p^2}{q^2}$
$\Rightarrow \frac{\frac{p}{2}[2a_1+(p-1\left)d\right]}{\frac{q}{2}[2a_1+(q-1\left)d\right]}=\frac{p^2}{q^2}\Rightarrow q[2a_1+(p-1\left)d\right]=p[2a_1+(q-1\left)d\right]\Rightarrow 2a_1(q-p)=d(q-p)\Rightarrow 2a_1=d(\because p\ne q)$
$\therefore \frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}=\frac{a_1+10a_1}{a_1+40a_1}=\frac{11}{41}$