Difference between revisions of "2015 AMC 8 Problems/Problem 18"
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==Problem== | ==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 <math>5\times5</math> array is an arithmetic sequence with five terms | + | 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 <math>5\times5</math> array is an arithmetic sequence with five terms. What is the value of <math>\text{X}</math>? |
<math>\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42</math> | <math>\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42</math> |
Revision as of 23:15, 19 November 2020
Contents
[hide]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 ?
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.