Arithmetic Series

Question

Three consecutive terms of an arithmetic sequence have a sum of 12 and a product of -80. Find the terms. Let the terms be x-d, x, and x+d.

 

Answer 1

(x-d) + x + (x+d) = 3x = 12, so x = 4

 

(4-d) * 4 * (4+d) = -80

 

64 - 4d^2 = -80

 

-4d^2 = -144

 

d^2 = 36

 

d = +/- 6

The three numbers are -2, 4, and 10.  The product is -80.

 

 

Answer 2

Let the terms be x-d, x, and x+d.

 

"Sum of 12"   means     (x-d) + (x) + (x+d) = 12

                                    3x = 12

                                     x = 4

       The middle term is 4.  Now, let's determine d:

             (4-d)(4)(4+d) = -80

             (4)(4-d)(4+d) = -80          [rearrange, because (a-b)(a+b) is a²-b²]

              4(16-d²) = -80

                16 - d² = -20              [divide both sides by 4)

                     - d² = -36             [subtract 16 from both sides]

                      d² = 36               [multiply by (-1)]

                     d = ±6

 

  The sequence may be:

          -2,  4,  10  ...             [sum 12; product (-80)]

      or

          10,  4.   -2  ...            [sum 12, product (-80)]