Mathematics - Sequence & Series Question with Solution | TestHub

$a=25,d=22-25=-3\text{. }$ Let $n$ be the no. of terms $\begin{array}{l}\text{ Sum }=116;\\ \text{ Sum }=\frac{n}{2}[2a+(n-1)d]\\ 116=\frac{n}{2}[50+(n-1)(-3)]\\ \text{ or }232=n[50-3n+3]=n[53-3n]=-3n^2+53n\\ \Rightarrow 3n^2-53+232=0\\ \Rightarrow (n-8)(3n-29)=0\\ \Rightarrow n=8\text{ or }n=\frac{29}{3},\\ n\ne \frac{29}{3}\\ \therefore n=8\\ \therefore \text{ Now, }{\mathrm{T}}_8=a+(8-1)d=25+7\times (-3)=25-21=4\\ \therefore \text{ Last term }=4\end{array}$