Answer
a. $a_{n} = \frac{5}{2}(-\frac{1}{2})^{n-1}$
b. 4th term: -$\frac{5}{16}$
Work Step by Step
a. $a_{n} = a(r)^{n-1}$ is the formula for the nth term of a geometric sequence. In this case a = $\frac{5}{2}$ and r=-$\frac{1}{2}$. Plug these values into the formula to get the formula for the nth term of this sequence.
b. To find the fourth term plug in n = 4 into the sequence: $a_{n} = \frac{5}{2}(-\frac{1}{2})^{4-1} = \frac{5}{2}(-\frac{1}{2})^{3} = -\frac{5}{16}$
