The sum of the first n terms of an A.P. is $3n^2 + 6n$. Find the nth term of this A.P.

Given:

The sum of first $n$ terms of an A.P. is $3n^{2} +6n$.

To do:

We have to find the $n^{th}$ term of the given A.P.

Solution:

$S_{n} =3n^{2} +6n$

For $n=1,\ S_{1} =3\times 1^{2} +6\times 1=3+6=9$

Therefore, first term $a=9$

For $n=2,\ S_{2} =3\times 2^{2} \ +6\times 2=12+12=24$

$\therefore$ Second term of the A.P.$=S_{2} -S_{1}$

$=24-9$

$=15$

Common difference of the A.P., $d=$second term $-$ first term

$=15-9=6$

We know that,

$a_{n}=a+(n-1)d$

$\therefore a_n=9+( n-1) \times 6$

$=9+6n-6$

$=6n+3$

Therefore, the $n^{th}$ term of the given A.P. is $6n+3$.

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Updated on: 10-Oct-2022

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