Question

Let $a_1, a_2, a_3,...$ be an $A.P$, such that $\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+a_{3}+\ldots+a_{q}} = \frac{p^{3}}{q^{3}} ; p\ne q$ Then $\frac{a_{6}}{a_{21}}$ is equal to:

• A. $\frac{41}{11}$
• B. $\frac{31}{121}$
• C. $\frac{11}{41}$
• D. $\frac{121}{1861}$

Answer 1

Answer: (B)

$\frac{31}{121}$

Answer 2

$\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+a_{3}+\ldots+a_{q}} = \frac{p^{3}}{q^{3}}$
$\Rightarrow\quad \frac{a_{1}+a_{2}}{a_{1}} = \frac{8}{1} \Rightarrow a_{1}+\left(a_{1}+d\right) = 8a_{1}$
$\Rightarrow\quad d=6a_{1}$
Now $\frac{a_{6}}{a_{21}} = \frac{a_{1}+5d}{a_{1}+20d}$
$= \frac{a_{1}+5\times6a_{1}}{a_{1}+20\times 6a_{1}} = \frac{1+30}{1+120} = \frac{31}{121}$