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Difference between revisions of "1976 AHSME Problems/Problem 4"

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The sum of a geometric series with <math>n</math> terms, initial term <math>a</math>, and ratio <math>r</math> is <math>\frac{a(1-r^n)}{1-r}</math>. So, <math>s=\frac{(1-r^n)}{1-r}</math>. Our initial sequence is <math>1, r, r^2, \sdots, r^n</math>, and replacing each terms with its reciprocal gives us the sequence <math>1, \frac{1}{r}, \frac{1}{r^2}, \sdots, \frac{1}{r^n}</math>. The sum is now <math>\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)}</math>.~MathJams
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The sum of a geometric series with <math>n</math> terms, initial term <math>a</math>, and ratio <math>r</math> is <math>\frac{a(1-r^n)}{1-r}</math>. So, <math>s=\frac{(1-r^n)}{1-r}</math>. Our initial sequence is <math>1, r, r^2, \dots, r^n</math>, and replacing each terms with its reciprocal gives us the sequence <math>1, \frac{1}{r}, \frac{1}{r^2}, \dots, \frac{1}{r^n}</math>. The sum is now <math>\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)}</math>.~MathJams
 
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}
 
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}

Latest revision as of 19:02, 12 July 2020

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

1976 AHSME (Problems • Answer Key • Resources)
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