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?

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) (Show Source): You can put this solution on YOUR website!

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) and --> d=2; a=1 --> 1, 3, 5, 7, ...

OR

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

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