Let a1 , a2 , a3 ,… be an A.P, such that 3 1 2 3 123 … ; … + ++ = ¹ + + ++ p q aa a p p q aa a a q . Then 6 21 a a is equal to:

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