Answer :

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

Sum of first n terms = Sn = 3n2 + 5n


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


an = (3n2 + 5n) - [3(n - 1)2 + 5(n - 1)]


= (3n2 + 5n) - [3(n2 + 1 - 2n) + 5n - 5]


= 3n2 + 5n - 3n2 - 3 + 6n - 5n + 5


= 2 + 6n


Now, common difference = d = an - an - 1


= 2 + 6n - [2 + 6(n - 1)]


= 2 + 6n - 2 - 6n + 6


= 6


Common difference = 6


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


Sum of first n terms = Sn = 3n2 + 5n


Put n = 1, we get S1 = 8


Put n = 2, we get S2 = 22


Now S1 = a1


a2 = S2 - S1


a2 = 22 - 8 = 14


Now, d = a2 - a1


= 14 - 8


= 6


Common difference = 6


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