Difference between revisions of "2015 AMC 8 Problems/Problem 18"
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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: | 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: | ||
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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>. | 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=== |
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>. | ||
− | ==Solution 3== | + | ===Solution 3=== |
The value of <math>X</math> is simply the average of the average values of both diagonals that contain <math>X</math>. This is <math>\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}</math> | The value of <math>X</math> is simply the average of the average values of both diagonals that contain <math>X</math>. This is <math>\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{\textbf{(B)}~31}</math> |
Revision as of 16:43, 16 January 2021
Problem
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. Each row and each column in this array is an arithmetic sequence with five terms. What is the value of ?
Solutions
Solution 1
We begin filling in the table. The top row has a first term and a fifth term , 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 and a fifth term of , 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 and a fifth term of . 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 .
Solution 3
The value of is simply the average of the average values of both diagonals that contain . This 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.