Given that \(-6, -2\frac{1}{2}, ..., 71\) is a linear sequence , calculate the number of terms in the sequence. 

A.

20

B.

21

C.

22

D.

23

Correct answer is D

\(T_{n} = a + (n - 1)d\) (for a linear or arithmetic progression)

Given: \(T_{n} = 71, a = -6, d = -2\frac{1}{2} - (-6) = 3\frac{1}{2}\)

\(\implies 71 = -6 + (n - 1)\times 3\frac{1}{2}\)

\(71 = -6 + 3\frac{1}{2}n - 3\frac{1}{2} = -9\frac{1}{2} + 3\frac{1}{2}n\)

\(71 + 9\frac{1}{2} = 3\frac{1}{2}n  \implies  n = \frac{80\frac{1}{2}}{3\frac{1}{2}}\)

\(= 23\)