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$Let a_{1}, a_{2}, a_{3} \dots be an A.P such that \frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{3}}{q^{3}}, p \neq q then \frac{a_{6}}{a_{21}} is$

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Correct option is B

take p=2,q=1 then the given equation becomes= a1+a2a1=8⇒a1+a1+da1=8⇒d=6a1 Now a6a21=a1+5da1+20d=a1+5×6a1a1+20×6a1=1+301+120=31121 n'th term in A.P=a+n-1d

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Let a1,a2,a3… be an A.P such that a1+a2+…+apa1+a2+…+aq=p3q3,p≠q then a6a21 is

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$Let a_{1}, a_{2}, a_{3} \dots be an A.P such that \frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{3}}{q^{3}}, p \neq q then \frac{a_{6}}{a_{21}} is$