In an A.P., the sum of first n terms is
Question:

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

Solution:

Here, the sum of first n terms is given by the expression,

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

We need to find the 25th term of the A.P.

So we know that the nthterm of an A.P. is given by,

$a_{n}=S_{n}-S_{n-1}$

So $a_{25}=S_{25}-S_{24} \ldots \ldots$ (1)

So, using the expression given for the sum of n terms, we find the sum of 25 terms (S25) and the sum of 24 terms (S24). We get,

$S_{25}=\frac{3(25)^{2}}{2}+\frac{13}{2}(25)$

$=\frac{3(625)}{2}+\frac{13(25)}{2}$

$=\frac{1875}{2}+\frac{325}{2}$

$=\frac{2200}{2}$

$=1100$

Similarly,

$S_{24}=\frac{3(24)^{2}}{2}+\frac{13}{2}(24)$

$=\frac{3(576)}{2}+\frac{13(24)}{2}$

$=\frac{1728}{2}+\frac{312}{2}$

$=\frac{2040}{2}$

$=1020$

Now, using the above values in (1),

$a_{25}=S_{25}-S_{24}$

$=1100-1020$

 

$=80$

Therefore, $a_{25}=80$.

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