SOLUTION: Find 3 numbers in the geometrical progression whose sum is 28 and whose product is 512

Question 1066281: Find 3 numbers in the geometrical progression whose sum is 28 and whose product is 512
Answer by ikleyn(52855) (Show Source): You can put this solution on YOUR website!
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Find 3 numbers in the geometric progression whose sum is 28 and whose product is 512
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Let the middle term be "x" and the common ratio be r.

Then the first of the three terms is , and the third term is .


The product of the three terms is  = .

Thus  = 512, which implies x = 8.

So, you just found the middle term. It is 8.


Then the sum of the three terms is  = 28, which gives you an equation

 = 28 - 8,   or   = 28 - 8 = 20,   or    = ,   or   =  = 2.5

which has the roots  r = 2  and/or  r= .


Thus the progression is  {4, 8, 16}   or   {16, 8, 4}.

Solved.

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