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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
Let a1,a2,a3,…{a_1},{a_2},{a_3}, \ldots be terms of an AP. If a1+a2+…+apa1+a2+…+aq=p2q2,  p≠q\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},\;p \ne q, then a6a21\frac{{{a_6}}}{{{a_{21}}}} equals
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Correct option: C
1141\frac{{11}}{{41}}
Option C is correct because by using the given ratio of sums and properties of arithmetic progressions, we can find the ratio of the 6th term to the 21st term.

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