In an A.P., the sum of the first $\displaystyle n$ terms is given by $\displaystyle S_{n}=6 n-n^{2}$. Find its $\displaystyle 30^{\text {th }}$ term.

Here \(\displaystyle S_{n}=6 n-n^{2}\) \[\mathrm{n}=1, \mathrm{~S}_{1}=\mathrm{a}=5 \] \(\displaystyle n=2, a+(a+d)=12-4\) or \(\displaystyle 2 a+d=8\) Putting \(\displaystyle \mathbf{a}=\mathbf{5}, \mathbf{d}=-\mathbf{2}\) Hence \(\displaystyle \mathbf{a}_{\mathbf{3 0}}=\mathbf{5}+\mathbf{2 9}(-\mathbf{2})=-\mathbf{5 3}\)