Difference between revisions of "1953 AHSME Problems/Problem 25"
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| + | ==Problem== | ||
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In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is: | In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is: | ||
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad \textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{2}{\sqrt{5}}</math> | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad \textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{2}{\sqrt{5}}</math> | ||
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| + | ==Solution== | ||
Given first term <math>a</math> and common ratio <math>r</math>, we have <math>a=a*r+a*r^2</math>, and. We divide by <math>a</math> in the first equation to get <math>1=r+r^2</math>. Rewriting, we have <math>r^2+r-1=0</math>. We use the quadratic formula to get <math>r = \frac{-1+-\sqrt{1^2-4(1)(-1)}}{2(1)}</math>. Because the terms all have to be positive, we must add the discriminant, getting an answer of <math>\frac{\sqrt{5}-1}{2}</math> <math>\boxed{C}</math>. | Given first term <math>a</math> and common ratio <math>r</math>, we have <math>a=a*r+a*r^2</math>, and. We divide by <math>a</math> in the first equation to get <math>1=r+r^2</math>. Rewriting, we have <math>r^2+r-1=0</math>. We use the quadratic formula to get <math>r = \frac{-1+-\sqrt{1^2-4(1)(-1)}}{2(1)}</math>. Because the terms all have to be positive, we must add the discriminant, getting an answer of <math>\frac{\sqrt{5}-1}{2}</math> <math>\boxed{C}</math>. | ||
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| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1953|num-b=24|num-a=26}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 23:54, 25 January 2020
Problem
In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:
Solution
Given first term 
and common ratio 
, we have 
, and. We divide by in the first equation to get 
. Rewriting, we have 
. We use the quadratic formula to get 
. Because the terms all have to be positive, we must add the discriminant, getting an answer of 
.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
