Question

five consecutive terms of an arithmetic sequence have a sum of 40. The product of the middle and the two end terms is 224. Find the terms of the sequence.

Answer 1

Let the five terms be: a, a + d, a + 2d, a + 3d, a + 4d, then
a + a + d + a + 2d + a + 3d + a + 4d = 5a + 15d = 40
i.e. a + 3d = 8
Also, (a + 2d)(a + 3d)(a + 4d) = 224
(a + 3d - d)(a + 3d)(a + 3d + d) = 224
(8 - d)(8)(8 + d) = 224
(8 - d)(8 + d) = 224/8 = 28
64 - d^2 = 28
d^2 = 64 - 28 = 36
d = sqrt(36) = 6
But a + 3d = 8
a + 3(6) = 8
a = 8 - 18 = -10

Therefore, the term of the sequence is: -10, -10 + 6, -10 + 2(6), -10 + 3(6), -10 + 4(6)
= -10, -4, -10 + 12, -10 + 18, -10 + 24
= -10, -4, 2, 8, 14