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

Given $A.P.25,22,19,...$ and sum $=116$
Here $a=25,d=22-25=-3$ ad $S_n=116$
As we know,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\frac{n}{2}[2\times 25+(n-1\left)\right(-3\left)\right]=116$
$\Rightarrow n(50-(n-1\left)\right(3\left)\right]=232$
$\Rightarrow n[50-3n+3]=232$
$\Rightarrow n[53-3n]=232$
$\Rightarrow 53n-3n^2=232$
$\Rightarrow 3n^2-53n+232=0$
$\Rightarrow 3n^2-24n-29n+232=0$
$\Rightarrow 3n(n-8)-29(n-8)=0$
$\Rightarrow (n-8)(3n-39)=0$
$\Rightarrow n=8,\frac{29}{3}$
$\because n\in N$
$\therefore n=8$
Thus, $8^{th}$ term is the last term of the A.P.

Finding last term
$a_n=a+(n-1)d$
Substitute $n=8$
$a_8=25+(8-1)(-3)$
$\Rightarrow a_8=25-7\times 3$
$\therefore a_8=25-21=4$

Final answer: Hence, the last term of given $A.P$ is $4$