Question 656233: The sum of three consecutive terms of an arithmetic sequence is 3. The sum of their squares is 75. find the three numbers
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The sum of three consecutive terms of an arithmetic sequence is 3. The sum of their squares is 75. find the three numbers
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Let a = the 2nd number in the sequence
Since an arithmetic sequence has a common difference, d, the 2nd number will a + d
And the 3rd number will be (a+d) + d = a + 2d
The sum of the three numbers is 3:
a + (a+d) + (a+2d) = 3
3a + 3d = 3
a + d = 1
a = 1 - d
The sum of squares is 75:
a^2 + (a+d)^2 + (a+2d)^2 = 75
Substitute a = 1-d:
(1-d)^2 + ((1-d)+d)^2 + ((1-d)+2d)^2 = 75
Solve for d:
1 + 2d + d^2 + 1 + 1 - 2d + d^2 = 75
3 + 2d^2 = 75
d^2 = 36
This gives d = 6, d = -6
Let's take d=6 [taking d=-6 gives the same 3 numbers-- check this]
So the 1st number a = 1 - d = -5
The 2nd number is -5 + 6 = 1
And the 3rd number is -5 + 2*6 = 7
Ans: -5,1,7
Check:
-5+1+7 = 3
25+1+49 = 75
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