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Question

The sum of first n terms of an AP is 3n2n. Find the 25th term of this AP.

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Solution

Given Sn=3n2n

Then Sn1=3[3(n1)]2(n1)

an=SnSn1

an=[3n2n][3(n1)2(n1)]

=3[n2(n1)2]n+n1

=3(n2n2+2n1)1=3(2n1)1=6n4

Then, a25=6×254=1504=146

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