smc

Question

In the five-sided star shown, the letters A, B, C, D and E are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the numbers at the ends of the line segments AB, BC, CD, DE, and EA form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence? FIGURE

Accepted Answer

Let the terms in the arithmetic sequence be a, a + d, a + 2d, a + 3d, and a + 4d. We seek the middle term a + 2d. These five terms are A + B, B + C, C + D, D + E, and E + A, in some order. The numbers A, B, C, D, and E are equal to 3, 5, 6, 7, and 9, in some order, so A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30. Hence, the sum of the five terms is (A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60. But adding all five numbers, we also get a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d, so 5a + 10d = 60. Dividing both sides by 5, we get a + 2d = 12, which is the middle term. The answer is (D).