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Difference between revisions of "1962 AHSME Problems/Problem 14"
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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  
 
Now, in this geometric series, <math>a=4</math>, and <math>r=-\frac23</math>. Plugging these into the formula, we get  
 
<math>\frac4{1-(-\frac23)}</math>, which simplifies to <math>\frac{12}5</math>, or <math>\boxed{2.4\textbf{ (B)}}</math>.
 
<math>\frac4{1-(-\frac23)}</math>, which simplifies to <math>\frac{12}5</math>, or <math>\boxed{2.4\textbf{ (B)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 13|num-a=15}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:16, 3 October 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 $\boxed{2.4\textbf{ (B)}}$.

See Also

1962 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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