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$Let a_{1}, a_{2} \dots a_{n} be the terms of an A.P. \frac{a_{1} + a_{2} + \dots a_{p}}{a_{1} + a_{2} + \dots + a_{1}} = \frac{p^{2}}{q^{2}}, p \neq q . T h e n \frac{a_{6}}{a_{21}} =$

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Correct option is A

p22a1+(p−1)dq22a1+(q−1)d=p2q2⇒2a1+(p−1)d2a1+(q−1)d=pq⇒a1+p−12da1+q−12d=pq For a6a21,p=11,q=41⇒a6a21=1141 where p−12=5 p=11q−12=20 q=41

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Let a1,a2…an be the terms of an A.P. a1+a2+…apa1+a2+…+a1=p2q2,p≠q. Then a6a21=

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$Let a_{1}, a_{2} \dots a_{n} be the terms of an A.P. \frac{a_{1} + a_{2} + \dots a_{p}}{a_{1} + a_{2} + \dots + a_{1}} = \frac{p^{2}}{q^{2}}, p \neq q . T h e n \frac{a_{6}}{a_{21}} =$