\[\boxed{\text{698}\text{\ (698)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ a_{n} = 2n + 1\]
\[a_{1} = 3,\ \ S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{2n + 1 + 3}{2} \cdot n = n(n + 2);\]
\[\textbf{б)}\ a_{n} = 3 - n,\ \ a_{1} = 2,\ \]
\[S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{2 + 3 - n}{2} \cdot n = \frac{n(5 - n)}{2}.\]