The sum of the first n terms of an A.P. is $4n^2 + 2n$. Find the nth term of this A.P.

Given:

The sum of first $n$ terms of an A.P. is $4n^{2} +2n$.

To do:

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

Solution:

$S_{n} =4n^{2} +2n$

For $n=1,\ S_{1} =4\times 1^{2} +2\times 1=4+2=6$

Therefore, first term $a=6$

For $n=2,\ S_{2} =4\times 2^{2} \ +2\times 2=16+4=20$

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

$=20-6$

$=14$

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

$=14-6=8$

We know that,

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

$\therefore a_n=6+( n-1) \times 8$

$=6+8n-8$

$=8n-2$

Therefore, the $n^{th}$ term of the given A.P. is $8n-2$. 

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Updated on: 10-Oct-2022

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