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JAMB Mathematics 2017 Paper

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Question : 15

Total: 44

Find the sum to infinity of the series

\frac{1}{4}

\frac{1}{8}

\frac{1}{16}

,..........

$\frac{1}{2}$
$\frac{3}{5}$
$\frac{-1}{5}$
$\frac{73}{12}$

Solution:

Concept:

The sum to infinity of a geometric progression is

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

when

|r| < 1

.

Explanation:

Identify the first term:

a = \frac{1}{4}

.

Find the common ratio:

\displaystyle r = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{8} \times \frac{4}{1} = \frac{1}{2}

.

Since

r = \frac{1}{2}

is less than

1

in magnitude, the sum to infinity exists.

Apply the formula:

\displaystyle S_\infty = \frac{a}{1 - r} = \frac{\frac{1}{4}}{1 - \frac{1}{2}}

.

Simplify:

\displaystyle S_\infty = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2}

.

Answer:

\frac{1}{2}

(Option A).

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