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

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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