Each term beyond the first is -2 times the previous one, and there are n-1 terms beyond the first, so the n'th term is given by (-2)n-1 times the first term or: f(n) = (-2)n-1*(-3.5)
The fourth term is then: f(4) = (-2)3*(-3.5) = -8*(-3.5) = 28
Each term beyond the first is -2 times the previous one, and there are n-1 terms beyond the first, so the n'th term is given by (-2)n-1 times the first term or: f(n) = (-2)n-1*(-3.5)
The fourth term is then: f(4) = (-2)3*(-3.5) = -8*(-3.5) = 28