In an A.P., the sum of first $n$ terms is $\frac{3n^2}{2}+\frac{13}{2}n$. Find its 25th term.

Given:

In an A.P., the sum of first $n$ terms is $\frac{3n^2}{2}+\frac{13}{2}n$.

To do:

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

Solution:

$S_{n} =\frac{3n^2}{2}+\frac{13}{2}n$

For $n=1,\ S_{1} =\frac{3(1)^2}{2}+\frac{13}{2}(1)=\frac{3+13}{2}=\frac{16}{2}=8$

Therefore, first term $a=8$

For $n=2,\ S_{2} =\frac{3(2)^2}{2}+\frac{13}{2}(2)=\frac{12+26}{2}=\frac{38}{2}=19$

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

$=19-8$

$=11$

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

$=11-8=3$

We know that,

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

$\therefore a_{25}=8+( 25-1) \times 3$

$=8+24\times 3$

$=8+72$

$=80$

Therefore, the $25^{th}$ term of the given A.P. is $80$. 

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Updated on: 10-Oct-2022

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