Question

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

Answer 1

we have,

A.P. is

$25,22,\mathrm{19......}$

$S_n=116$

$a=25$

$d=T_2-T_1$

$d=22-25$

$d=-3$

Then, we know that

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

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

$\Rightarrow 116=\frac{n}{2}(50-3n+3)$

$\Rightarrow 116=\frac{n}{2}(53-3n)$

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

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

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

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

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

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

If

$3n-8=0$

$n=\frac{8}{3}\left(\text{not}\text{possible}\right)$

If

$n-8=0$

$n=8$

So,

The last term is

$S_n=\frac{n}{2}(a+l)$

$116=\frac{8}{2}(25+l)$

$116=4(25+l)$

$116=100+4l$

$116-100=4l$

$4l=16$

$l=4$

Hence, the last term of this series is

4.

This is the answer.