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JAMB Mathematics 2024 Paper

Question : 14

Total: 50

\;\frac{1}{2}

Solution:

Concept:

The sum to infinity of a geometric progression (GP) is given by

S_\infty = \frac{a}{1 - r}

, valid when the common ratio

|r| < 1

Explanation:

We are given

S_\infty = 100

and

r = -\frac{1}{2}

Substitute these values into the formula:

100 = \frac{a}{1 - \left(-\frac{1}{2}\right)}

Simplify the denominator:

1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}

Therefore,

100 = \frac{a}{\frac{3}{2}}

Multiply both sides by

\frac{3}{2}

to solve for

a

a = 100 \times \frac{3}{2} = 150

Thus, the first term is 150.

Answer:

A. 150

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