Difference between revisions of "2015 AMC 8 Problems/Problem 18"
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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>. | 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>. | ||
| − | + | ==Solution 2== | |
| − | The middle term of the first row is <math>\frac{25+1}{2}=13</math>, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is <math>\frac{17+81}{2}=49</math>. Applying this again for the middle column, the answer is <math>\frac{49+13}{2}=\boxed{\textbf{(B)~31}</math>. | + | The middle term of the first row is <math>\frac{25+1}{2}=13</math>, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is <math>\frac{17+81}{2}=49</math>. Applying this again for the middle column, the answer is <math>\frac{49+13}{2}=\boxed{\textbf{(B)}~31}</math>. |
==See Also== | ==See Also== | ||
Revision as of 17:37, 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 
?
![[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((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]](http://latex.artofproblemsolving.com/a/5/e/a5e463abe580eaf777a86e44ccfbdd1acb90f714.png)
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:
![[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]](http://latex.artofproblemsolving.com/c/0/2/c028f3edd8159a97e7b756a43ead831d4eb02ad3.png)
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:
![[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]](http://latex.artofproblemsolving.com/5/4/0/540608620fbcae9b59ab2dc17ec5eab91393e103.png)
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 
.
Solution 2
The middle term of the first row is 
, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is 
. Applying this again for the middle column, the answer 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. 
