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Question

Let a1,a2,a3,... be the terms of an AP. If a1+a2+a3+...+apa1+a2+a3+...+aq=p2q2, pq, find the value of a6a21.

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Solution

Given, a1,a2,a3,... are in AP.

and a1+a2+a3+...+apa1+a2+a3+...+aq=p2q2

p/2[2a1+(p1)d]q/2[2a1+(q1)d]=p2q2

[ Sn=n2{2a+(n1)d}]

2a1+(p1)d2a1+(q1)d=pq

2a1q+pqdqd=2a1p+pqdpd

2a1qqd2a1p+pd=0

2a1 p2a1qpd+qd=0

2a1(pq)d(pq)=0

(2a1d)(pq)=0

2a1d=0 [ pq]

d=2a1

Now, a6a21=a1+5da1+20d=a1+10a1a1+40a1=1141


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