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 11:38, 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 | |
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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