\(\mathrm{a}=3, \mathrm{~T}_5=\mathrm{a}+4 \mathrm{~d}=9\). Then,
\(3+4 d=9,4 d=9-3=6, d=6 / 4=3 / 2\)
\(\mathrm{S}_{\mathrm{n}}=81, \mathrm{a}=3, \mathrm{~d}=3 / 2\)
\(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\)
\(81=\frac{n}{2}\left[2(3)+(n-1) \frac{3}{2}\right]\)
\(81=\frac{n}{2}\left[6+\frac{3 n-3}{2}\right]\)
\(81=\frac{n}{2}\left[\frac{12+3 n-3}{2}\right]\)
\(81=\frac{n}{2}\left[\frac{9+3 n}{2}\right]\)
\(81 \times 4=n(9+3 n)\)
\(324=9 \mathrm{n}+3 \mathrm{n}^2\)ivide through by 3
\(108=3 n+n^2\)
then,
\(n^2+3 n-108=0\)
\(n^2+12 n-9 n-108=0\)
\(n(n+12)-9(n+12)=0\)
\begin{array}{l}
(n-9)(n+12)=0 \\
n=9 \text { or }-12
\end{array}s \(n\) cannot be negative, \(n=9\)