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Difference between revisions of "1962 AHSME Problems/Problem 14"

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

Revision as of 19:55, 16 April 2014

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 $\boxed{2.4\textbf{ (B)}}$.

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