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
waec 2016furthermathematics

The correct answer is D. 23

\(T_{n} = a + (n - 1)d\) (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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