Question

Let a₁, a₂, a₃, .... be an A.P, such that $\frac{a_{1}+a_{2}+...+a_{p}}{a_{1}+a_{2}+a_{3}+...+a_{q}}=\frac{p^{3}}{q^{3}};\ p\neq q.\ Then\ \frac{a_{6}}{a_{21}}$ is equal to:
(a) $\frac{41}{11}$
(b) $\frac{31}{121}$
(c) $\frac{11}{41}$
(d) $\frac{121}{1861}$

Answer 1

option C (mistake p^2/q^2)