Expert: Steve Holleran - 9/30/2007

Question
the book asks
"find the first eight numbers in an arithmetic sequence in which the sum of the first and seventh terms is 40 and the product of the first and forth term is 160."

our teacher told us that the book said the answers were (8,12,16,20,24,28,32,36) but i have no idea how to do it.

Answer
Hi Matthew,

Okay, let's remember that an arithmetic sequence has a common difference, d, between terms, and that we can find the "n th" term using the formula:

                a(n) = a(1) + (n - 1) *d

So, then  the 7th term would be :  a(7) = a(1) + 6d

and the 4th term would bea(4) = a(1) + 3d.

If the sum of the first and seventh is 40, we have:

     a(1) + [a(1) + 6d] = 40

or           2a(1) = 40 - 6d

or            a(1) = 20 - 3d

Then if the product of the first and fourth is 160, you have

             a(1) * [a(1) + 3d] = 160

substituting for a(1),

             (20 - 3d) * (20 - 3d + 3d) = 160

             (20 - 3d) * (20) = 160

             (20 - 3d) = 8  so a(1) = 8.

Going back to the first equation we got, substitute 8 for a(1):

              a(1) + [a(1) + 6d] = 40

               8   + [ 8  + 6d] = 40

                        16 + 6d = 40

                             6d = 24   so d = 4.

Then you have your sequence :  8,12,16,...


Steve Holleran