Difference between revisions of "2003 AMC 10B Problems/Problem 24"
(→Video Solution by SpreadTheMtahLove) |
|||
(22 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | The | + | The first four terms in an arithmetic sequence are <math>x+y</math>, <math>x-y</math>, <math>xy</math>, and <math>\frac{x}{y}</math>, in that order. What is the fifth term? |
<math> \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} </math> | <math> \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} </math> | ||
− | + | ==Solution 1== | |
− | ==Solution== | ||
The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing. | The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing. | ||
Line 19: | Line 18: | ||
Substitute into our other equation. | Substitute into our other equation. | ||
− | <cmath> | + | <cmath> |
− | \frac{x}{y} | + | \frac{x}{y}=x-5y</cmath> |
− | \frac{-3}{y-1} | + | <cmath>\frac{-3}{y-1}=\frac{-3y}{y-1}-5y</cmath> |
− | -3 | + | <cmath>-3=-3y-5y(y-1)</cmath> |
− | 0 | + | <cmath>0=5y^2-2y-3</cmath> |
− | 0 | + | <cmath>0=(5y+3)(y-1)</cmath> |
− | y | + | <cmath>y=-\frac35, 1</cmath> |
− | But <math>y</math> cannot be <math>1</math> because then | + | But <math>y</math> cannot be <math>1</math> because then the first term would be <math>x+1</math> and the second term <math>x-1</math> while the last two terms would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally, |
<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath> | <cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Because this is an arithmetic sequence, we conclude from the first two terms that the common difference is <math>-2y</math>. Therefore, <math>xy = x-3y</math> and <math>\frac{x}{y}=x-5y</math>. If we multiply <math>xy</math> and <math>\frac{x}{y}</math>, we see: | ||
+ | |||
+ | <math>xy\cdot\frac{x}{y} = (x-3y)(x-5y) = x^2 - 8xy + 15y^2</math> | ||
+ | |||
+ | Because <math>xy\cdot\frac{x}{y}</math>, by basic multiplication, is <math>x^2</math>, we have | ||
+ | |||
+ | <math>x^2 = x^2 - 8xy + 15y^2</math> | ||
+ | |||
+ | <math>8xy = 15y^2</math> | ||
+ | |||
+ | <math>8x = 15y</math> | ||
+ | |||
+ | <math>x = \frac{15y}{8}</math> | ||
+ | |||
+ | Now that we have <math>x</math> in terms of <math>y</math>, we substitute in <math>\frac{15y}{8}</math> in for <math>x</math> in <math>\frac{x}{y}</math> (the fourth term). This leaves us with <math>\frac{\frac{15y}{8}}{y} = \frac{15}{8}</math>. | ||
+ | |||
+ | Recall that <math>\frac{x}{y}</math> can be written as <math>x - 5y</math>. Thus, <math>x - 5y = \frac{15}{8}</math>. Substitute in <math>\frac{15}{8}</math> in for <math>x</math>, and we see: | ||
+ | |||
+ | <math>\frac{15y}{8} - 5y = \frac{15}{8}</math> | ||
+ | |||
+ | <math>15y - 40y = 15</math> | ||
+ | |||
+ | <math>y = \frac{15}{-25} = \frac{-3}{5}</math> | ||
+ | |||
+ | Aha! This means <math>2y</math>, the common difference, is <math>2\cdot y = 2\cdot \frac{-3}{5} = \frac{-6}{5}</math>. Now, all we need to do is find the fifth term, which is just <math>\frac{x}{y}-2y</math>. We can substitute known values to solve: | ||
+ | |||
+ | <math>\frac{x}{y}-2y = \frac{15}{8} - \frac{-6}{5} = \frac{15}{8} + \frac{6}{5} = \frac{75 + 48}{40} = \boxed{\textbf{(E)}\ \frac{123}{40}}</math>. | ||
+ | |||
+ | ~SXWang | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | From the first two terms, we can figure out the common difference, <math>x-y-(x+y)=-2y</math>. | ||
+ | This means that the third, fourth and fifth terms are <math>x-3y</math>, <math>x-5y</math> and <math>x-7y</math> respectively. | ||
+ | |||
+ | The third term is also equal to <math>xy</math>, so | ||
+ | <math>xy=x-3y</math>, which can be rearranged to <math>xy+3y=x</math>. | ||
+ | |||
+ | Dividing by y, we have | ||
+ | <math>x+3=\frac{x}{y}</math>. | ||
+ | |||
+ | <math>\frac{x}{y}</math> happens to be the fourth term. Therefore, | ||
+ | <math>\frac{x}{y}=x-5y=x+3</math>. | ||
+ | |||
+ | <math>x-5y=x+3</math> | ||
+ | |||
+ | <math>-5y=3</math> | ||
+ | |||
+ | <math>y=-\frac{3}{5}</math> | ||
+ | |||
+ | Now that we have <math>y</math>, we can substitute it into an equation above, like <math>\frac{x}{y}=x+3</math>. | ||
+ | |||
+ | <math>\frac{x}{\frac{-3}{5}}=x+3</math> | ||
+ | |||
+ | <math>-\frac{5x}{3}=x+3</math> | ||
+ | |||
+ | <math>-\frac{5x+3x}{3}=3</math> | ||
+ | |||
+ | <math>-\frac{8x}{3}=3</math> | ||
+ | |||
+ | <math>8x=-9</math> | ||
+ | |||
+ | <math>x=\frac{-9}{8}</math> | ||
+ | |||
+ | As mentioned earlier, the fifth term we want in the end is equal to <math>x-7y</math>. Substitute some more and... | ||
+ | |||
+ | <math>\frac{-9}{8}-7(\frac{-3}{5})=\frac{-9\cdot5}{8\cdot5}+\frac{-3\cdot-7\cdot8}{5\cdot8}=\frac{-45}{40}+\frac{168}{40}=\boxed{\textbf{(E)}\ \frac{123}{40}}</math> | ||
+ | |||
+ | (Alternatively, like the above solutions suggest, you could also use the admittedly easier <math>\frac{x}{y}-2y</math> as the final step.) | ||
+ | |||
+ | ~ a seesaw named owlly81 ~ | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=wzNnrj51BAQ | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:45, 1 October 2024
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
Because this is an arithmetic sequence, we conclude from the first two terms that the common difference is . Therefore, and . If we multiply and , we see:
Because , by basic multiplication, is , we have
Now that we have in terms of , we substitute in in for in (the fourth term). This leaves us with .
Recall that can be written as . Thus, . Substitute in in for , and we see:
Aha! This means , the common difference, is . Now, all we need to do is find the fifth term, which is just . We can substitute known values to solve:
.
~SXWang
Solution 3
From the first two terms, we can figure out the common difference, . This means that the third, fourth and fifth terms are , and respectively.
The third term is also equal to , so , which can be rearranged to .
Dividing by y, we have .
happens to be the fourth term. Therefore, .
Now that we have , we can substitute it into an equation above, like .
As mentioned earlier, the fifth term we want in the end is equal to . Substitute some more and...
(Alternatively, like the above solutions suggest, you could also use the admittedly easier as the final step.)
~ a seesaw named owlly81 ~
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=wzNnrj51BAQ
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.