Let $a_1, a_2, a_3, ....$ cannot be terms of an AP. If $\frac{a_1+a_2+....+a_p}{a_1 + a_2 + ....+ a_q} = \frac{p^2}{q^2}, \; p \neq q$, then $\frac{a_6}{a_{21}}$ equals :

Question

( A ) $\frac{7}{2}$
( B ) $\frac{11}{41}$
( C ) $\frac{2}{7}$
( D ) $\frac{41}{11}$