# The sum of first n terms of an AP is (4n2 + 2n). The nth term of this AP isA. (6n – 2)B. (7n – 3)C. (8n – 2)D. (8n + 2)

Let Sn denotes the sum of first n terms of an AP.

Sum of first n terms = Sn = 4n2 + 2n

Then nth term is given by: an = Sn - Sn - 1

an = (4n2 + 2n) - [4(n - 1)2 + 2(n - 1)]

= (4n2 + 2n) - [4(n2 + 1 - 2n) + 2n - 2]

= 4n2 + 2n - 4n2 - 4 + 8n - 2n + 2

= 8n - 2

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