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

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

Revision as of 11:24, 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 $\boxed{\textbf{(A)}}$.

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