Question

(i) The sum of first n terms of an AP is $\left(\frac{5n^2}{2}+\frac{3n}{2}\right)$. Find the nth term and the 20th term of this AP. [CBSE 2006C]

(ii) The sum of the first n terms of an AP is $\left(\frac{3n^2}{2}+\frac{5n}{2}\right)$. Find its nth term and the 25th term. [CBSE 2006C]

Answer 1

(i)
$$\begin{gathered} s_n=\frac{5n^2}{2}+\frac{3n}{2} \\ \mathrm{Sum}\mathrm{of}1\mathrm{term}=5\left(\frac{1}{2}\right)+3\left(\frac{1}{2}\right)=4 \\ \mathrm{Sum}\mathrm{of}2\mathrm{term}s=5\left(\frac{4}{2}\right)+3\left(\frac{2}{2}\right)=13 \\ 2\mathrm{nd}\mathrm{term}=\mathrm{Sum}\mathrm{of}\mathrm{first}2\mathrm{terms}-\mathrm{Sum}\mathrm{of}1\mathrm{term}=13-4 \\ =9 \\ \mathrm{Common}\mathrm{difference},d=9-4=5 \\ n\mathrm{th}\mathrm{term}\mathrm{i}.\mathrm{e}{a}_n=4+\left(n-1\right)5 \\ =5n-1 \\ 20\mathrm{th}\mathrm{term}\mathrm{i}.\mathrm{e}{a}_{20}=4+\left(20-1\right)5 \\ =99 \end{gathered}$$


(ii) Let S_n denotes the sum of first n terms of the AP.
$$\begin{gathered} \therefore S_n=\frac{3n^2}{2}+\frac{5n}{2} \\ \Rightarrow S_{n-1}=\frac{3{\left(n-1\right)}^2}{2}+\frac{5\left(n-1\right)}{2} \\ =\frac{3\left(n^2-2n+1\right)}{2}+\frac{5\left(n-1\right)}{2} \\ =\frac{3n^2-n-2}{2} \end{gathered}$$

∴ n^th term of the AP, a_{n
$$\begin{gathered} =S_n-S_{n-1} \\ =\left(\frac{3n^2+5n}{2}\right)-\left(\frac{3n^2-n-2}{2}\right) \\ =\frac{6n+2}{2} \\ =3n+1 \end{gathered}$$}

Putting n = 25, we get

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

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