Q. 524.7( 6 Votes )

The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.

Answer :

We have sum of n terms, Sn = 3n2 + 4n

Put n = 1


S1 = T1 = 3(1)2 + 4 (1) = 7


Put n = 2


S2 = 3(2)2 + 4 (2) = 20


T2 = S2 – S1 = 20 – 7 = 13


Put n = 3


S3 = 3(3)2 + 4(3) = 39


T3 = S3 – S2 = 39 – 20 = 19


Therefore, first term is 7 and common difference, d = 13 – 7 = 6


The 25th term is, Tn= a + (n – 1) d


T25 = 7 + (25 – 1) * 6


= 7 + 24 * 6 = 151


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