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.

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