document.write( "Question 390900: if a,b,c are in continued proportion, prove that:
\n" ); document.write( "(a^2+b^2+c^3)/(a+b+c)=a-b+c\r
\n" ); document.write( "\n" ); document.write( "and hence find three numbers in continued proportion so that their sum is 14 and the sum of their squares is 84
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Algebra.Com's Answer #277237 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
You definitely mean $\"%28a%5E2%2Bb%5E2%2Bc%5E2%29%2F%28a%2Bb%2Bc%29=a-b%2Bc\"$ .\r
\n" ); document.write( "\n" ); document.write( "The given is $\"a%2Fb+=+b%2Fc\"$ , that is, a,b,c are in continued proportion . One consequence of this condition is the fact that $\"ac+=+b%5E2\"$ , which we will use later.\r
\n" ); document.write( "\n" ); document.write( "To start off, consider $\"%28a%2Bb%2Bc%29%28a-b+%2B+c%29\"$ . Upon expansion, this is equal to

.
\n" ); document.write( "But $\"ac+=+b%5E2\"$ , so after substitution,
\n" ); document.write( " $\"%28a%2Bb%2Bc%29%28a+-b+%2B+c%29+=+a%5E2+%2B+b%5E2+%2B+c%5E2\"$ . Assuming that the sum a+b+c is not equal to zero, then we divide by a+b+c, and the result follows.\r
\n" ); document.write( "\n" ); document.write( "Consider a = 8, b = 4, and c = 2. Then $\"%2864%2B16+%2B+4%29%2F%288%2B4%2B2%29+=+6\"$ . \n" ); document.write( "

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