Difference between revisions of "2015 AMC 8 Problems/Problem 18"
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− | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term | + | ==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. 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> | ||
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</asy> | </asy> | ||
− | ==Solution== | + | ==Solutions== |
− | 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: | + | |
+ | ===Solution 1=== | ||
+ | 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> | ||
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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> | ||
− | 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=== | ||
− | 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=== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/tKsYSBdeVuw?t=1258 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
==See Also== | ==See Also== | ||
Latest revision as of 23:09, 27 January 2021
Contents
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
Video Solution
https://youtu.be/tKsYSBdeVuw?t=1258
~ pi_is_3.14
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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