SOLUTION: If the sum of first four terms of an arithmetic sequence is 16 and their product 105 find the terms if they all are integers ... I know that the A.s is (1,3,5,7) but how to get it

Algebra ->  Sequences-and-series -> SOLUTION: If the sum of first four terms of an arithmetic sequence is 16 and their product 105 find the terms if they all are integers ... I know that the A.s is (1,3,5,7) but how to get it      Log On


   

Question 955519: If the sum of first four terms of an arithmetic sequence is 16 and their product 105 find the terms if they all are integers ... I know that the A.s is (1,3,5,7) but how to get it
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
If the first term is "a" and the difference is "d", then the first four terms 
are 

a, a+d, a+2d, a+3d

the sum of first four terms of an arithmetic sequence is 16 
So, 

a + a+d + a+2d + a+3d = 16
              4a + 6d = 16 

Divide thru by 2 

              2a + 3d = 8

their product 105, 
So  
   a(a+d)(a+2d)(a+3d) = 105 

So we have this system of two equations: 

system%282a+%2B+3d+=+8%2Ca%28a%2Bd%29%28a%2B2d%29%28a%2B3d%29+=+105%29 

Solve the first equation for "a":

2a = 8-3d

 a = %288-3d%29%2F2

Substitute in the second equation:

a%28a%2Bd%29%28a%2B2d%29%28a%2B3d%29+=+105



Get LCD in each parentheses:



Combine fractions in parentheses:



%28%288-3d%29%2F2%29%28%288-d%29%2F2%29%28%288%2Bd%29%2F2%29%28%288%2B3d%29%2F2%29+=+105

Multiply both sides by 16

%288-3d%29%288-d%29%288%2Bd%29%288%2B3d%29+=+1680

Rearrange the factors on the left:

%288-3d%29%288%2B3d%29%288-d%29%288%2Bd%29+=+1680

FOIL

%2864-9d%5E2%29%2864-d%5E2%29+=+1680

FOIL again:

4096-640d%5E2%2B9d%5E4=1680

Get 0 on the right side:

2416-640d%5E2-9d%5E4=0

Get in descending order:

9d%5E4-640d%5E2%2B2416=0

Factor the trinomial:

%289d%5E2-604%29%28d%5E2-4%29=0

Factor further:

%289d%5E2-604%29%28d-2%29%28d%2B2%29=0

9d%5E2-604=0, d-2=0, d%2B2=0
9d%5E2=604, d=2, d=-2
d%5E2=604%2F9
d=+%22%22+%2B-+sqrt%28604%2F9%29
d=+%22%22+%2B-+sqrt%284%2A151%29%2F3
d=+%22%22+%2B-+2sqrt%28151%29%2F3

We can ignore the irrational values for d, since the sequence 
has only integers.

The only integer values for d are d=2 and d=-2

For d=2, substitute in

 a = %288-3d%29%2F2
 a = %288-3%2A2%29%2F2
 a = %288-6%29%2F2
 a = 2%2F2
 a = 1

Substitute d=2 and a=1 in

a, a+d, a+2d, a+3d

1, 1+2, 1+2(2), 1+3(2)

Therefore 1 solution is

1, 3, 5, 7

----------------------------

For d=-2, substitute in

 a = %288-3d%29%2F2
 a = %288-3%2A%28-2%29%29%2F2
 a = %288%2B6%29%2F2
 a = 14%2F2
 a = 7

Substitute d=-2 and a=7 in

a, a+d, a+2d, a+3d

7, 7+(-2), 7+2(-2), 7+3(-2)

7, 5, 7-4, 7-6

Therefore another solution is

7, 5, 3, 1.

But this is the same sequence reversed.

Edwin