Given that \(-6, -2\frac{1}{2}, ..., 71\) is a linear sequence , calculate the number of terms in the sequence.
\(T_{n} = a + (n - 1)d\) (for a linear or arithmetic progression)
Given: \(T_{n} = 71, a = -6, d = -2\frac{1}{2} - (-6) = 3\frac{1}{2}\)
\(\implies 71 = -6 + (n - 1)\times 3\frac{1}{2}\)
\(71 = -6 + 3\frac{1}{2}n - 3\frac{1}{2} = -9\frac{1}{2} + 3\frac{1}{2}n\)
\(71 + 9\frac{1}{2} = 3\frac{1}{2}n \implies n = \frac{80\frac{1}{2}}{3\frac{1}{2}}\)
\(= 23\)
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