$\begin{array}{1 1} 7 \\ 1 \\ 4 \\ -2 \end{array} $

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- Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[(2a+(n-1)d]$
- $n^{th}\:term=t_n=a+(n-1)d$

Given that sum of $n$ terms of the $A.P.$ is $116$

$i.e.,\:\:S_n= 25+22+19+........+t_n=116$

In this A.P. first term $a=25,$ common difference $d=22-25=-3$

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

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

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

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

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

$\Rightarrow\:n=8\:\:or\:\:n=\large\frac{29}{3}$

But since the number of terms $n$ cannot be fraction, $n=8$

$\therefore \:$ The last term $t_n=a+(n-1)d$

$=25+(8-1)(-3)=4$

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