SOLUTION: if a,b,c are in continued proportion, prove that: (a^2+b^2+c^3)/(a+b+c)=a-b+c and hence find three numbers in continued proportion so that their sum is 14 and the sum of their

Question 390900: if a,b,c are in continued proportion, prove that:
(a^2+b^2+c^3)/(a+b+c)=a-b+c
and hence find three numbers in continued proportion so that their sum is 14 and the sum of their squares is 84

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
You definitely mean .
The given is , that is, a,b,c are in continued proportion . One consequence of this condition is the fact that , which we will use later.
To start off, consider . Upon expansion, this is equal to .
But , so after substitution,
. Assuming that the sum a+b+c is not equal to zero, then we divide by a+b+c, and the result follows.
Consider a = 8, b = 4, and c = 2. Then .
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