Question

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

• A. $\frac{4}{34}$
• B. $\frac{9}{37}$
• C. $\frac{11}{41}$
• D. $\frac{15}{46}$

Answer 1

Answer: (C)

$\frac{11}{41}$

Let, First term of an A.P be $a$ and common difference be $d$ then,

Sum of $n$ terms of an A.P $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Given that,

$\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3...+a_q}=\frac{p^2}{q^2}$

$⟹\frac{\frac{p}{2}[2a+(p-1\left)d\right]}{\frac{q}{2}[2a+(q-1\left)d\right]}=\frac{p^2}{q^2}$

$⟹\frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p}{q}$

Put, $p=2p-1$ and $q=2q-1$ we get,

$⟹\frac{2a+(2p-2)d}{2a+(2q-2)d}=\frac{2p-1}{2q-1}$

$⟹\frac{a+(p-1)d}{a+(q-1)d}=\frac{2p-1}{2q-1}$

$⟹\frac{a_p}{a_q}=\frac{2p-1}{2q-1}$

$\therefore \frac{a_6}{a_{21}}=\frac{2\left(6\right)-1}{2\left(21\right)-1}=\frac{11}{41}$ (Ans)