The sum of the first n terms of an AP is $3 n^{2} + 6 n$ . Find the n th term of this AP.

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Solution

Given
$S_{n} = 3 n^{2} + 6 n$
$S_{n} = \frac{n}{2} \times (a + l)$
$3 n^{2} + 6 n = \frac{n}{2} \times (a + l)$
l = 6n+12-a

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The sum of the first n terms of an AP is 3n2+6n. Find the n th term of this AP.

Given Sn=3n2+6n Sn=n2×(a+l) 3n2+6n=n2×(a+l) l = 6n+12-a

The sum of the first n terms of an AP is $3 n^{2} + 6 n$ . Find the n th term of this AP.

Given
$S_{n} = 3 n^{2} + 6 n$
$S_{n} = \frac{n}{2} \times (a + l)$
$3 n^{2} + 6 n = \frac{n}{2} \times (a + l)$
l = 6n+12-a