Question

Let a₁, a₂, a₃............ be terms on A.P. If $\frac{a_1 + a_2 + ......... a_p}{a_1 + a_2 + ......... a_q} = \frac{p^2}{q^2}$, p ≠ q, then $\frac{a_6}{a_{21}}$ equals
$(a) \frac{41}{11} \\(b) \frac{7}{2} \\(c) \frac{2}{7} \\(d) \frac{11}{41}$

Answer 1

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