Let $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$ where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

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

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