Sum

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.

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#### Solution

We have,

S_{n} = `1/2`[3n^{2} + 7n]

Replacing n by (n - 1), we get

S_{n - 1} = `1/2`[3(n - 1)^{2} + 7(n - 1)]

⇒ S_{n - 1} = `1/2`[3(n^{2} + 1 - 2n) + 7n - 7]

⇒ S_{n - 1} = `1/2`[3n^{2} + 3 - 6n + 7n - 7]

⇒ S_{n - 1} = `1/2`[3n^{2} + n - 4]

Now, nth term = S_{n} - S_{n - 1}

⇒ a_{n} = `1/2[3n^2 + 7n] - 1/2[3n^2 + n - 4]`

⇒ a_{n} = `1/2`[3n^{2} + 7n - 3n^{2} - n + 4]

⇒ a_{n} = `1/2[6n + 4]`

⇒ a_{n} = 3n + 2

Now,

a_{20 }= 3 × 20 + 2 = 60 + 2 = 62.

Concept: Sum of First n Terms of an AP

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