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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a
4111
b
31121
c
1141
d
1211861
answer is B.
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Detailed Solution
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