Answer :

Let an be the nth term of the AP.

To find: an and a25


Since, an =Sn - Sn - 1


= ( ) - ()


= 1/2 (3n2 + 5n) - 1/2 [3(n - 1)2 + 5(n - 1)]


= 1/2 (3n2 + 5n - 3n2 - 3 + 6n - 5n + 5)


= 1/2 (6n - 2)


= 3n + 1


Since an = 5n - 1


For 25th term, put n = 25, we get,


a25 = 3(25) + 1


= 75 + 1


= 76


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