# Difference between revisions of "1951 AHSME Problems/Problem 11"

## Problem

The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 - r}$ where $a$ denotes the first term and $- 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is: $\textbf{(A)}\ \frac {a^2}{(1 - r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 - r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 + r^2} \qquad\textbf{(E)}\ \text{none of these}$

## Solution

Let the original geometric series be $a,ar,ar^2,ar^3,ar^4\cdots$. Therefore, their squares are $a^2,a^2r^2,a^2r^4,a^2r^6,\cdots$, which is a geometric sequence with first term $a^2$ and common ratio $r^2$. Thus, the sum is $\boxed{\textbf{(C)}\ \frac {a^2}{1 - r^2}}$.

## See Also

 1951 AHSC (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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