Let $a_{1},a_{2},a_{3},\ldots$ be terms of an AP. If $\frac{a_{1} + a_{2} + \ldots + a_{p}}{a_{1} + a_{2} + \ldots + a_{q}} = \frac{p^{2}}{q^{2}},\ p
eq q$, then $\frac{a_{6}}{a_{21}}$ equals

Question

Let $a_{1},a_{2},a_{3},\ldots$ be terms of an AP. If $\frac{a_{1} + a_{2} + \ldots + a_{p}}{a_{1} + a_{2} + \ldots + a_{q}} = \frac{p^{2}}{q^{2}},\ p 
eq q$, then $\frac{a_{6}}{a_{21}}$ equals

Dr. Abubacker · Accepted Answer

(c)
Given, $\frac{\frac{p}{2}\lbrack 2a_{1} + (p - 1)d\rbrack}{\frac{q}{2}\lbrack 2a_{1} + (q - 1)d\rbrack} = \frac{p^{2}}{q^{2}}$
$\Rightarrow \ \frac{\left( 2a_{1} - d \right) + pd}{\left( 2a_{1} - d \right) + qd} = \frac{p}{q}$
$\Rightarrow \ \left( 2a_{1} - d \right)(p - q) = 0$
$\Rightarrow \ a_{1} = \frac{d}{2}$
Now, $\frac{a_{6}}{a_{21}} = \frac{a_{1} + 5d}{a_{2} + 20d}$
$= \frac{\frac{d}{2} + 5d}{\frac{d}{2} + 20d} = \frac{11}{41}$

Q. Let $a_{1},a_{2},a_{3},\ldots$ be terms of an AP. If $\frac{a_{1} + a_{2} + \ldots + a_{p}}{a_{1} + a_{2} + \ldots + a_{q}} = \frac{p^{2}}{q^{2}},\ p \neq q$, then $\frac{a_{6}}{a_{21}}$ equals

Explanation

Reviewed by Dr. Abubacker

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