2015 AMC 8

Complete problem set with solutions and individual problem pages

Problem 18 Hard

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. The square in the center is labelled X as shown. What is the value of X?

  • A.

    21

  • B.

    31

  • C.

    36

  • D.

    40

  • E.

    42

Answer:B

Solution 1

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:

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:

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}.