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

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