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, \ | + | 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}, \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 |
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Revision as of 19:02, 12 July 2020
Problem 4
Let a geometric progression with n terms have first term one, common ratio 
and sum 
, where and are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is
Solution
The sum of a geometric series with 
terms, initial term 
, and ratio is 
. So, 
. Our initial sequence is 
, and replacing each terms with its reciprocal gives us the sequence $1, \frac{1}{r}, \frac{1}{r^2}, \sdots, \frac{1}{r^n}$ (Error compiling LaTeX. Unknown error_msg). The sum is now $\frac{1-(\frac{1}{r^n})}{1-\frac{1}{r}=\frac{\frac{(1-r^n)}{r^n}}{\frac{1-r}{r}}=\frac{s}{r^{n-1}}\Rightarrow \textbf{(C)}$ (Error compiling LaTeX. Unknown error_msg).~MathJams
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