Question

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

Answer 1

$\frac{a_1+a_2+....+a_p}{a_1+a_2+....+a_q}=\frac{p^2}{q^2}$

$\Rightarrow \frac{\frac{p}{2}[2a_1+(p-1)d]}{\frac{q}{2}[2a_1+(q-1)d]}=\frac{p^2}{q^2}$

$\Rightarrow \frac{2a_1+(p-1)d}{2a_1+(q-1)d}=\frac{p}{q}$ by cancelling of the same terms both sides

$\Rightarrow 2a_{1}q+(p-1)qd=2a_{1}p+(q-1)pd$ by cross multiplication

$\Rightarrow 2a_1(q-p)=[(q-1)p-(p-1)q]d$ by grouping the common terms

$\Rightarrow -2a_1(p-q)=[pq-p-pq+q]d$

$\Rightarrow -2a_1(p-q)=-(p-q)d$ on simplifying

$\Rightarrow 2a_1=d$.....$\left(1\right)$

So $\frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}$

$=\frac{a_1+5\left(2a_1\right)}{a_1+20\left(2a_1\right)}$ From $\left(1\right)$

$=\frac{11a_1}{41a_1}=\frac{11}{41}$