SOLUTION: Find five consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156.

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Question 1012889: Find five consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156.
Found 2 solutions by fractalier, ikleyn:
Answer by fractalier(6550)   (Show Source): You can put this solution on YOUR website!
The five terms are a, a+d, a+2d, a+3d and a+4d.
If we add them we get
5a + 10d = 115 or
a + 2d = 23 or
a = 23 - 2d
We also have
(a+d)^2 + (a+3d)^2 = 1156
We can substitute from before into this and get
(23 - 2d + d)^2 + (23 - 2d + 3d)^2 = 1156
(23 - d)^2 + (23 + d)^2 = 1156
529 - 46d + d^2 + 529 + 46d + d^2 = 1156
2d^2 + 1058 = 1156
2d^2 = 98
d^2 = 49
d = 7, -7
a is then 23 - 2(7) = 9 or 23 - 2(-7) = 37
Thus our two possibilities appear to be
9, 16, 23, 30, 37 and
37, 30, 23, 16, 9

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
Find five consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156.
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You can write these terms in this way

a-2d, a-d a, a+d, a+2d

where a if the third (the middle of 5) term, and d is the common difference.

Then the sum of 5 terms is 5a, and you easily find a =  = 23.

To find d, use the second part of the condition

 = 1156,   or

 =  = 578,   or

 = 578 -  = 49.

Hence, d = +/-7. 

Although you have two solutions for d, the progression is actually unique.

It is 9, 16, 23, 30, 37.


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