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