The sum of first n terms of an A.P. is $3n^2 + 4n$. Find the 25th term of this A.P.

Given:

The sum of first $n$ terms of an A.P. is $3n^{2} +4n$.

To do:

We have to find the $25^{th}$ term of the given A.P.

Solution:

$S_{n} =3n^{2} +4n$

For $n=1,\ S_{1} =3\times 1^{2} +4\times 1=3+4=7$

Therefore, first term $a=7$

For $n=2,\ S_{2} =3\times 2^{2} \ +4\times 2=12+8=20$

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

$=20-7$

$=13$

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

$=13-7=6$

We know that,

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

$a_{25}=7+( 25-1) \times 6$

$=7+24\times 6$

$=7+144$

$=151$

Therefore, the $25^{th}$ term of the given A.P. is 151.

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Updated on: 10-Oct-2022

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