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

• A. 41/11
• B. 7/2
• C. 2/7
• D. 11/41

Answer 1

Answer: (D)

11/41

Given $a_1,a_2,a_3,....$ be terms of A.P

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

$\Rightarrow \frac{\frac{p}{2}[2a_1+(p-1)d]}{\frac{q}{2}[2a_1+(q-1)d]}=\frac{p^2}{q^2}$ $[\because (S_n)=\frac{n}{2}[2a+(n-1\left)d\right]]$

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

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

$\Rightarrow 2a_1(q-p)=d[(q-1)p-(p-1)q]$

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

$\Rightarrow \frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}=\frac{a_1+10a_1}{a_1+40a_1}$

$\Rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}$