Difference between revisions of "1953 AHSME Problems/Problem 25"
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==Solution== | ==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,asdf getting an answer of <math>\frac{\sqrt{5}-1}{2}</math> <math>\boxed{C}</math>. |
==See Also== | ==See Also== |
Revision as of 10:38, 16 May 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,asdf getting an answer of .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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All AHSME Problems and Solutions |
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