SOLUTION: a^2+3ad+2d^2=15, 2a+ 3d=8 Two possible progressions a = 1, d = 2 1,3,5,7 or a = 7, d = -2 7,5,3,1.....how?

Algebra ->  Sequences-and-series -> SOLUTION: a^2+3ad+2d^2=15, 2a+ 3d=8 Two possible progressions a = 1, d = 2 1,3,5,7 or a = 7, d = -2 7,5,3,1.....how?      Log On


   

Question 1154148: a^2+3ad+2d^2=15,
2a+ 3d=8
Two possible progressions
a = 1, d = 2
1,3,5,7
or
a = 7, d = -2
7,5,3,1.....how?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


a%5E2%2B3ad%2B2d%5E2+=+%28a%2Bd%29%28a%2B2d%29+=+15

2a%2B3d+=+%28a%2Bd%29%28a%2B2d%29+=+8

So (a+d) and (a+2d) are two numbers whose product is 15 and whose sum is 8.

The two numbers are, in some order, 3 and 5.

So either

(a) a%2Bd=3 and a%2B2d=5 --> d=2; a=1 --> 1, 3, 5, 7, ...

OR

(b) a%2Bd=5 and a%2B2d=3 --> d=-2; a=7 --> 7, 5, 3, 1, ...