# 1976 AHSME Problems/Problem 4

## Problem 4

Let a geometric progression with n terms have first term one, common ratio $r$ and sum $s$, where $r$ and $s$ are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is $\textbf{(A) }\frac{1}{s}\qquad \textbf{(B) }\frac{1}{r^ns}\qquad \textbf{(C) }\frac{s}{r^{n-1}}\qquad \textbf{(D) }\frac{r^n}{s}\qquad \textbf{(E) } \frac{r^{n-1}}{s}$

## Solution

The sum of a geometric series with $n$ terms, initial term $a$, and ratio $r$ is $\frac{a(1-r^n)}{1-r}$. So, $s=\frac{(1-r^n)}{1-r}$. Our initial sequence is $1, r, r^2, \dots, r^n$, and replacing each terms with its reciprocal gives us the sequence $1, \frac{1}{r}, \frac{1}{r^2}, \dots, \frac{1}{r^n}$. The sum is now $\frac{1-(\frac{1}{r^n})}{1-\frac{1}{r}}=\frac{\frac{(1-r^n)}{r^n}}{\frac{1-r}{r}}=\frac{s}{r^{n-1}}\Rightarrow \textbf{(C)}$.~MathJams

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