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Difference between revisions of "2015 AMC 8 Problems/Problem 18"
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==Solution==
 
==Solution==
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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:
+
We begin filling in the table.  The top row has a first term <math>1</math> and a fifth term <math>25</math>, 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>
 
<asy>
 
size(3.85cm);
 
size(3.85cm);
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</asy>
 
</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:
+
The fifth row has a first term of <math>17</math> and a fifth term of <math>81</math>, 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>
 
<asy>
 
size(3.85cm);
 
size(3.85cm);
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</asy>
 
</asy>
  
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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 <math>13</math> and a fifth term of <math>49</math>.  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==
 
==Solution 2==

Revision as of 11:28, 30 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, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$. Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. What is the value of $X$?

$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$

[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]

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 $\frac{25-1}4=6$. 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 $\frac{81-17}4=16$. 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 $\frac{49-13}4=9$, so the third term is $13+2\cdot 9=\boxed{\textbf{(B) }31}$.

Solution 2

The middle term of the first row is $\frac{25+1}{2}=13$, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\frac{17+81}{2}=49$. Applying this again for the middle column, the answer is $\frac{49+13}{2}=\boxed{\textbf{(B)}~31}$.

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

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