# The sum of the first n terms of an AP is (3n2 + 6n). Find the nth term and the 15th term of this AP.

Given: The sum of the first n terms of an AP is (3n2 + 6n).

To find: the nth term and the 15th term of this AP.

Solution:

Sum of first n terms = Sn = 3n2 + 6n

Now let an be the nth term of the AP.

To find: an and a15

Since an =Sn - Sn - 1 ⇒ an= (3n2 + 6n) - (3(n - 1)2 + 6(n - 1))

⇒ an= (3n2 + 6n) - (3(n2 +1 - 2n) + 6(n - 1))

⇒ a= (3n2 + 6n) - (3n2 + 3 - 6n + 6n - 6)

⇒ a= 3n2 + 6n - 3n2 - 3 + 6n - 6n + 6

⇒ a= 6n + 3

Now, a15 = 6(15) + 3

⇒  a15= 93

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