Question

How many terms of the A.P. $25,22,19,...$ are needed to give the sum $116$? Also find the last term.

Answer 1

Here $a=25,d=-3,S_n=116$

we know

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$116=\frac{n}{2}[2\times 25+(n-1\left)\right(-3\left)\right]$

$\Rightarrow 232=n(50-3n+3)$

$\Rightarrow 232=n(53-3n)$

$\Rightarrow 232=53n-3n^2$

$\Rightarrow 3n^2-53n+232=0$

$\Rightarrow 3n^2-24n-29n+232=0$

$\Rightarrow 3n(n-8)-29(n-8)=0$

$\Rightarrow (3n-29)(n-8)=0$

$\therefore n=8,n\ne \frac{29}{3}$ (fraction)

$\therefore$ Number of terms required = '8'

$t_n=a+(n-1)d$

$\Rightarrow t_n=25-1(8-1)(-3)$

$\Rightarrow t_n=25-21=4$

Hence, Last term is '4'