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. $\frac{121}{1681}$

Answer 1

Answer: (D)

$ \frac{a_{1}+a_{2}+\ldots \ldots+a_{p}}{a_{1}+a_{2}+a_{3}+\ldots \ldots+a_{q}}=\frac{p^{3}}{q^{3}} $
$ \begin{aligned} \Rightarrow & \frac{\frac{p}{2}[2 a+(p-1) d]}{\frac{q}{2}[2 a+(q-1) d]}=\frac{p}{q^{3}} \\ \Rightarrow & \frac{2 a+(p-1) d}{2 a+(q-1) d}=\frac{p^{2}}{q^{2}} \end{aligned} $
putting $p =11$ and $q =41$
$ \Rightarrow \frac{2 a +10 d }{2 a +40 d }=\left(\frac{11}{41}\right)^{2} $
$ \Rightarrow \frac{ a +5 d }{ a +2 d }=\frac{121}{1681} $
$ \Rightarrow \frac{ a _{6}}{ a _{21}}=\frac{121}{1681} $