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Difference between revisions of "1959 AHSME Problems/Problem 12"

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Now we can calculate our geometric progression to be <math>20+25, 50+25, 100+25 = 45, 75, 125</math>. Therefore, the common ratio is <math>\frac{125}{75} = \frac{75}{45} = \boxed{\frac{5}{3}}</math>, and our answer is <math>\boxed{\textbf{(A)}}</math>.
 
Now we can calculate our geometric progression to be <math>20+25, 50+25, 100+25 = 45, 75, 125</math>. Therefore, the common ratio is <math>\frac{125}{75} = \frac{75}{45} = \boxed{\frac{5}{3}}</math>, and our answer is <math>\boxed{\textbf{(A)}}</math>.
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==See also==
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{{AHSME 50p box|year=1959|num-b=11|num-a=13}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 11:25, 21 July 2024

Problem

By adding the same constant to $20,50,100$ a geometric progression results. The common ratio is: $\textbf{(A)}\ \frac53 \qquad\textbf{(B)}\ \frac43\qquad\textbf{(C)}\ \frac32\qquad\textbf{(D)}\ \frac12\qquad\textbf{(E)}\ \frac{1}3$

Solution

Suppose that the constant is $x$. Then $20+x, 50+x, 100+x$ is a geometric progression, so $(20+x)(100+x) = (50+x)^2$. Expanding, we get $2000 + 120x + x^2 = 2500 + 100x + x^2$; therefore, $20x = 500$, so $x=25$.

Now we can calculate our geometric progression to be $20+25, 50+25, 100+25 = 45, 75, 125$. Therefore, the common ratio is $\frac{125}{75} = \frac{75}{45} = \boxed{\frac{5}{3}}$, and our answer is $\boxed{\textbf{(A)}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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