#### Question

If the sum of the first *n* terms of an A.P. is `1/2`(3n^{2} +7n), then find its *n*^{th} term. Hence write its 20^{th} term.

#### Solution

S_{n}=`1/2`(3n2+7n)

⇒S_{1}=12(3+7)=5

⇒S_{2}=12(3×2^{2}+7×2)=`26/2`=13

We know

S_{1}=a_{1}=5

S_{2}=a_{1}+a_{2}=13

⇒S_{2}−S_{1}=a_{1}+a_{2}−a_{1}

⇒13−5=a_{2}

⇒a_{2}=8

We know

d=a_{2}−a_{1}

⇒d=8−5=3

We know that the *n*th term of the AP is given by

a_{n}=a+(n−1)d

⇒a_{n}=5+(n−1)3

⇒a_{n}=2+3n

Thus, the *n*th term of the AP is 2+3n.

Thus, the 20th term is

a_{20}=2+3(20)=62

Hence, the 20th term of the AP is 62.

Is there an error in this question or solution?

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If the sum of the first n terms of an A.P. is 1/2 (3n^2 +7n), then find its nth term. Hence write its 20th term. Concept: Sum of First n Terms of an AP.

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