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Q.

$1 + (0.04) + {(0.04)}^{2} + \dots$ $\infty =$

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a

$\frac{25}{24}$

b

$\frac{24}{25}$

c

1

d

0.04

answer is C.

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Detailed Solution

$1 + (0.04) + {(0.04)}^{2} + \dots$

$a = 1, r = \frac{t_{2}}{t_{1}} = \frac{0.04}{1} = 0.04$

$s_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - 0.04} = \frac{1}{0.96} = \frac{100}{96}$

$= \frac{25}{24}$

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