Answer
(a) $430$
(b) $-\frac{1705}{128}$
Work Step by Step
(a) Given arithmetic sequence with $a_1=-20, d=14$, we have $a_{10}=(-20)+(10-1)(14)=106$ and $S_{10}=\frac{10}{2}(-20+106)=430$
(b) Given geometric sequence with $a_1=-20, r=-\frac{1}{2}$, we have $S_{10}=(-20)\times\frac{1-(-\frac{1}{2})^{10}}{1-(-\frac{1}{2})}=-\frac{1705}{128}$
