Difference between revisions of "2003 AMC 10B Problems/Problem 24"
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The common difference of the sequence is <math>(x-y)-(x+y)=-2y</math>. Then, the next two terms are equal to <math>x-y-2y=x-3y</math> and <math>x-3y-2y=x-5y</math>. We are also given that the next two terms are equal to <math>xy</math> and <math>x/y</math>, respectively, so we set up two equations: | The common difference of the sequence is <math>(x-y)-(x+y)=-2y</math>. Then, the next two terms are equal to <math>x-y-2y=x-3y</math> and <math>x-3y-2y=x-5y</math>. We are also given that the next two terms are equal to <math>xy</math> and <math>x/y</math>, respectively, so we set up two equations: | ||
− | <math>x-3y=xy</math> | + | <math>x-3y=xy</math> and |
<math>x-5y=x/y</math>. | <math>x-5y=x/y</math>. | ||
Then, | Then, | ||
<math>x=3y+xy \implies y=\frac{x}{x+3}</math>. | <math>x=3y+xy \implies y=\frac{x}{x+3}</math>. | ||
So, | So, | ||
− | <math>x-\frac{5x}{x+3}=x+3 \implies x(x+3)-5x=(x+3)^2</math> | + | <math>x-\frac{5x}{x+3}=x+3 \implies x(x+3)-5x=(x+3)^2</math>. |
We rearrange to get | We rearrange to get | ||
− | <math>x^2+3x-5x=x^2+6x+9 \implies 8x=-9 \implies x=-\frac{9}{8}</math> | + | <math>x^2+3x-5x=x^2+6x+9 \implies 8x=-9 \implies x=-\frac{9}{8}</math>. |
Then, | Then, | ||
− | <math>y=\frac{-\frac{9}{8}}{\frac{15}{8}}=-\frac{3}{5}</math> | + | <math>y=\frac{-\frac{9}{8}}{\frac{15}{8}}=-\frac{3}{5}</math>. |
So, the fifth term is | So, the fifth term is | ||
<math>x-7y=-\frac{9}{8}+\frac{21}{5}=\boxed{\frac{123}{40}}</math>. | <math>x-7y=-\frac{9}{8}+\frac{21}{5}=\boxed{\frac{123}{40}}</math>. |
Revision as of 17:00, 2 September 2020
Contents
[hide]Problem
The first four terms in an arithmetic sequence are , , , and , in that order. What is the fifth term?
Solution 1
The difference between consecutive terms is Therefore we can also express the third and fourth terms as and Then we can set them equal to and because they are the same thing.
Substitute into our other equation.
But cannot be because then the first term would be and the second term while the last two terms would be equal to Therefore Substituting the value for into any of the equations, we get Finally,
Solution 2
The common difference of the sequence is . Then, the next two terms are equal to and . We are also given that the next two terms are equal to and , respectively, so we set up two equations: and . Then, . So, . We rearrange to get . Then, . So, the fifth term is .
~peace09
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
All AMC 10 Problems and Solutions |
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