The sum of first n terms of an AP is
Question:

(i) The sum of first $n$ terms of an AP is $\left(\frac{5 n^{2}}{2}+\frac{3 n}{2}\right)$. Find the $n$th term and the 20 th term of this AP.

(ii) The sum of the first $n$ terms of an $\mathrm{AP}$ is $\left(\frac{3 n^{2}}{2}+\frac{5 n}{2}\right)$. Find its $n$th term and the 25 th term.

 

Solution:

(i)

$s_{n}=\frac{5 n^{2}}{2}+\frac{3 n}{2}$

Sum of 1 term $=5\left(\frac{1}{2}\right)+3\left(\frac{1}{2}\right)=4$

Sum of 2 term $s=5\left(\frac{4}{2}\right)+3\left(\frac{2}{2}\right)=13$

2 nd term $=$ Sum of first 2 terms $-$ Sum of 1 term $=13-4$

$=9$

Common difference, $d=9-4=5$

$n$th term i. e $a_{n}=4+(n-1) 5$

$=5 n-1$

20 th term i. e $a_{20}=4+(20-1) 5$

$=99$

(ii) Let Sn denotes the sum of first n terms of the AP.

$\therefore S_{n}=\frac{3 n^{2}}{2}+\frac{5 n}{2}$

$\Rightarrow S_{n-1}=\frac{3(n-1)^{2}}{2}+\frac{5(n-1)}{2}$

$=\frac{3\left(n^{2}-2 n+1\right)}{2}+\frac{5(n-1)}{2}$

$=\frac{3 n^{2}-n-2}{2}$

$\therefore n^{\text {th }}$ term of the $\mathrm{AP}, a_{n}$

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

$=\left(\frac{3 n^{2}+5 n}{2}\right)-\left(\frac{3 n^{2}-n-2}{2}\right)$

$=\frac{6 n+2}{2}$

$=3 n+1$

Putting = 25, we get

$a_{25}=3 \times 25+1=75+1=76$

Hence, the nth term is (3n + 1) and 25th term is 76.

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