# The sum of first n terms of an A.P. is given by Sn = 3n2 + 2n. Determine the A.P. and its 15th term.

Sn = 3n2 + 2n

Taking n = 1, we get

S1 = 3(1)2 + 2(1)

S1 = 3 + 2

S1 = 5

a1 = 5

Taking n = 2, we get

S2 = 3(2)2 + 2(2)

S2 = 12 + 4

S2 = 16

a2 = S2 – S1 = 16 – 5 = 11

Taking n = 3, we get

S3 = 3(3)2 + 2(3)

S3 = 27 + 6

S3 = 33

a3 = S3 – S2 = 33 – 16 = 17

So, a = 5,

d = a2 – a1 = 11 – 5 = 6

Now, we have to find the 15th term

an = a + (n – 1)d

a15 = 5 + (15 – 1)6

a15 = 5 + 14 × 6

a15 = 5 + 84

a15 = 89

Hence, the 15th term is 89 and AP is 5, 11, 17, 23,…

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