Difference between revisions of "2015 AMC 8 Problems/Problem 18"
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</asy> | </asy> | ||
+ | ==Solution== | ||
+ | We begin filling in the table. The top row has a first term 1 and a fifth term 25, so we have the common difference is <math>\frac{25-1}4=6</math>. This means we can fill in the first row of the table: | ||
+ | <asy> | ||
+ | size(3.85cm); | ||
+ | label("$X$",(2.5,2.1),N); | ||
+ | for (int i=0; i<=5; ++i) | ||
+ | draw((i,0)--(i,5), linewidth(.5)); | ||
+ | |||
+ | for (int j=0; j<=5; ++j) | ||
+ | draw((0,j)--(5,j), linewidth(.5)); | ||
+ | void draw_num(pair ll_corner, int num) | ||
+ | { | ||
+ | label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); | ||
+ | } | ||
+ | |||
+ | draw_num((0,0), 17); | ||
+ | draw_num((4, 0), 81); | ||
+ | draw_num((1,4), 7); | ||
+ | draw_num((2,4), 13); | ||
+ | draw_num((3,4), 19); | ||
+ | draw_num((0, 4), 1); | ||
+ | |||
+ | draw_num((4,4), 25); | ||
+ | |||
+ | |||
+ | void foo(int x, int y, string n) | ||
+ | { | ||
+ | label(n, (x+0.5,y+0.5), p = fontsize(19pt)); | ||
+ | } | ||
+ | |||
+ | foo(2, 4, " "); | ||
+ | foo(3, 4, " "); | ||
+ | foo(0, 3, " "); | ||
+ | foo(2, 3, " "); | ||
+ | foo(1, 2, " "); | ||
+ | foo(3, 2, " "); | ||
+ | foo(1, 1, " "); | ||
+ | foo(2, 1, " "); | ||
+ | foo(3, 1, " "); | ||
+ | foo(4, 1, " "); | ||
+ | foo(2, 0, " "); | ||
+ | foo(3, 0, " "); | ||
+ | foo(0, 1, " "); | ||
+ | foo(0, 2, " "); | ||
+ | foo(1, 0, " "); | ||
+ | foo(1, 3, " "); | ||
+ | foo(1, 4, " "); | ||
+ | foo(3, 3, " "); | ||
+ | foo(4, 2, " "); | ||
+ | foo(4, 3, " "); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | The fifth row has a first term of 17 and a fifth term of 81, so the common difference is <math>\frac{81-17}4=16</math>. We can fill in the fifth row of the table as shown: | ||
+ | <asy> | ||
+ | size(3.85cm); | ||
+ | label("$X$",(2.5,2.1),N); | ||
+ | for (int i=0; i<=5; ++i) | ||
+ | draw((i,0)--(i,5), linewidth(.5)); | ||
+ | |||
+ | for (int j=0; j<=5; ++j) | ||
+ | draw((0,j)--(5,j), linewidth(.5)); | ||
+ | void draw_num(pair ll_corner, int num) | ||
+ | { | ||
+ | label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); | ||
+ | } | ||
+ | |||
+ | draw_num((0,0), 17); | ||
+ | draw_num((4, 0), 81); | ||
+ | draw_num((1,4), 7); | ||
+ | draw_num((2,4), 13); | ||
+ | draw_num((3,4), 19); | ||
+ | draw_num((4, 4), 25); | ||
+ | draw_num((0, 4), 1); | ||
+ | draw_num((1, 0), 33); | ||
+ | draw_num((2, 0), 49); | ||
+ | draw_num((3, 0), 65); | ||
+ | |||
+ | |||
+ | |||
+ | void foo(int x, int y, string n) | ||
+ | { | ||
+ | label(n, (x+0.5,y+0.5), p = fontsize(19pt)); | ||
+ | } | ||
+ | |||
+ | foo(2, 4, " "); | ||
+ | foo(3, 4, " "); | ||
+ | foo(0, 3, " "); | ||
+ | foo(2, 3, " "); | ||
+ | foo(1, 2, " "); | ||
+ | foo(3, 2, " "); | ||
+ | foo(1, 1, " "); | ||
+ | foo(2, 1, " "); | ||
+ | foo(3, 1, " "); | ||
+ | foo(4, 1, " "); | ||
+ | foo(2, 0, " "); | ||
+ | foo(3, 0, " "); | ||
+ | foo(0, 1, " "); | ||
+ | foo(0, 2, " "); | ||
+ | foo(1, 0, " "); | ||
+ | foo(1, 3, " "); | ||
+ | foo(1, 4, " "); | ||
+ | foo(3, 3, " "); | ||
+ | foo(4, 2, " "); | ||
+ | foo(4, 3, " "); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | We must find the third term of the arithmetic sequence with a first term of 13 and a fifth term of 49. The common difference of this sequence is <math>\frac{49-13}4=9</math>, so the third term is <math>13+2\cdot 9=\boxed{\textbf{(B) }31}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=17|num-a=19}} | {{AMC8 box|year=2015|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:59, 25 November 2015
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, is an arithmetic sequence with five terms, in which the first term is and the constant added is . Each row and each column in this array is an arithmetic sequence with five terms. What is the value of ?
Solution
We begin filling in the table. The top row has a first term 1 and a fifth term 25, so we have the common difference is . This means we can fill in the first row of the table:
The fifth row has a first term of 17 and a fifth term of 81, so the common difference is . We can fill in the fifth row of the table as shown:
We must find the third term of the arithmetic sequence with a first term of 13 and a fifth term of 49. The common difference of this sequence is , so the third term is .
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.