Five consecutive terms of an arithmetic sequence have a sum of 40. the product of the first, middle and the last term is 224. find the terms of sequence

AP: a₁, a₂, a₃, a₄, a₅   (number of terms n = 5)

To refresh:
Last term in a AP :  a₅ = a₁ + (n-1)d  and S= (a₁ + a₅ )(n/2)
Given: (S=40) and  (a₁ x a₃ x a₅ = 224)
1) S = 40 = (a₁ +a₅)(5/2) ===> (a₁ + a₅)5 =80 and a₁ + a₅ = 16 (a)
2) Last term (a₅) = a₁ + (n-1)d, replace a₅ by its value in (a)
  a₁ + [a₁ + (n-1)d] = 16 ===> 2a₁ + (5-1)d =16 ===> a₁ + 2d = 8 (b)
3) (a₁)(a₃)(a₅) = 224 (c) , we know that a₃ = (a₁ +2 d) so a₃ =8 (as per b)
Rewrote (c): 
a₁ x 8 x a₅ = 224 or a₁ x a₅ = 28

4) Now we have  S = (a₁ + a₅) = 16    and  P= (a₁ x a₅) = 28

5) Find 2 numbers which sum are 16 & their product 28:

X² - S.X + P =0 Solve this quadratic equation & you will find :

a₁ = 14   and a₅ = 2
And from (b) yo'll find d,the common difference d = - 3
The AP IS THEN:
14, 11, 8, 5, 2
PROOF: S= 40 AND 14 x 8 x 2 =224