Question

If the sum of a certain number of terms of the A.P. 25, 22, 19, .... is 116, Find the last term.

Answer 1

Here a = 25, d = 22 -25 = -3.

Let sum of an terms be 116 then $S_n$ = 116

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

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

$\Rightarrow 232=n[50-3n+3]$

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

$I\therefore n=\frac{-(-53)\pm \sqrt{(-53{)}^2-4\times 3\times 232}}{2\times 3}$

= $\frac{53\pm \sqrt{2809-2784}}{6}$

= $\frac{53\pm \sqrt{2}5}{6}=\frac{53\pm 5}{6}$

$\therefore n=\frac{53\pm 5}{6}orn=\frac{53-5}{6}$

n = $\frac{58}{6}orn=\frac{48}{6}=8$

n = $\frac{58}{6}$ is not possible. Thus n = 8

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

$\therefore a_8=25+(8-1)\times -3=25-21=4$

Thus last term of the given A.P. is 4