SOLUTION: Find three consecutive terms of an arithmetic sequence, if the sum of the first and second terms is 30, and if the product of the first and third terms is 153.

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Question 1102616: Find three consecutive terms of an arithmetic sequence, if the sum of the first and second terms is 30, and if the product of the first and third terms is 153.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
First term = a-d
Second term = a
Third term = a+d

system%28%28a-d%29%2Ba=30%2C%28a-d%29%28a%2Bd%29=153%29

system%28a-d%2Ba=30%2Ca%5E2-d%5E2=153%29

system%282a-d=30%2Ca%5E2-d%5E2=153%29

Solve the first equation for d:

2a-30=d

Substitute 2a-30 for d in

a%5E2-d%5E2=153

a%5E2-%282a-30%29%5E2=153

a%5E2-%284a%5E2-120a%2B900%29=153

a%5E2-4a%5E2%2B120a-900=153

-3a%5E2%2B120a-900=153

Divide through by -3

a%5E2-40a%2B300=-51

Get 0 on the right:

a%5E2-40a%2B351=0

Factor:

%28a-13%29%28a-27%29=0

a-13=0 ;  a-27=0
   a=13;     a=27

Substitute each in

2a-30=d

2%2813%29-30=d;   2%2827%29-30=d
26-30=d;   54-30=d
-4=d;   24=d

So one solution is

First term = a-d = 13-(-4) = 13+4 = 17
Second term = a = 13
Third term = a+d = 13+(-4) = 13-4 = 9

17,13,9

And the other solution is

First term = a-d = 27-(24) = 3
Second term = a = 27
Third term = a+d = 27+(24) = 51

3,27,51

In both cases, the sum of the first and second 
terms is 30, and the product of the first and 
third terms is 153.

Edwin