1962 AHSME Problems/Problem 14

Problem

Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$ , as the number of terms increases without bound. Then $s$ equals:

$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$

Solution

The infinite sum of a geometric series with first term $a$ and common ratio $r$ ( $-1<r<1$ ) is $\frac{a}{1-r}$ . Now, in this geometric series, $a=4$ , and $r=-\frac23$ . Plugging these into the formula, we get $\frac4{1-(-\frac23)}$ , which simplifies to $\frac{12}5$ , or $\boxed{2.4\textbf{ (B)}}$ .

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1962 AHSME Problems/Problem 14

Problem

Solution

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