document.write( "Question 876239: If a^2,b^2,c^2 are in AP then prove that 1/b+c,1/c+a,1/a+b are also in AP. \n" ); document.write( "
Algebra.Com's Answer #528664 by Edwin McCravy(20060) $\"\"$ $\"About$
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If a², b², c² are in AP then prove that $\"1%2F%28b%2Bc%29\"$ , $\"1%2F%28c%2Ba%29\"$ , $\"1%2F%28a%2Bb%29\"$ are also in AP.
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document.write( "Since the second part is more complicated than the first part, we\r\n" );
document.write( "begin by proving the converse, and then we hope we can reverse the \r\n" );
document.write( "steps to prove the original.\r\n" );
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document.write( "So first we try to prove the converse:\r\n" );
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document.write( "If , ,  are in AP, then a², b², c² are also in AP.\r\n" );
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document.write( "Three terms are in AP, if and only if\r\n" );
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document.write( "Multiply through by the LCD of (c+a)(b+c)(a+b) assuming that LCD is not 0.\r\n" );
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document.write( "2(b+c)(a+b) = 1(c+a)(a+b) + 1(c+a)(b+c)\r\n" );
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document.write( "2(ba+b²+ca+cb) = (ca+cb+a²+ab) + (cb+c²+ab+ac)\r\n" );
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document.write( "2ba+2b²+2ca+2cb = ca+cb+a²+ab+cb+c²+ab+ac\r\n" );
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document.write( "Write product of letters in alphabetical order so we can\r\n" );
document.write( "easily identify like terms:\r\n" );
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document.write( "2ab+2b²+2ac+2bc = ac+bc+a²+ab+bc+c²+ab+ac\r\n" );
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document.write( "2ab+2b²+2ac+2bc = 2ac+2bc+a²+2ab+c²\r\n" );
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document.write( "Subtract 2ab+2ac+2bc from both sides:\r\n" );
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document.write( "2b² = a²+c²\r\n" );
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document.write( "Write 2b² as b²+b²\r\n" );
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document.write( "b²+b² = a²+c²\r\n" );
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document.write( "b²-a² = c²-b²\r\n" );
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document.write( "Therefore a²,b²,c² are in AP because\r\n" );
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document.write( "Three terms are in AP, if and only if\r\n" );
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document.write( "Now we must reverse the steps to prove the original theorem.\r\n" );
document.write( "We have only proved the converse.\r\n" );
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document.write( "We now assume that we are given that a²,b²,c² are in AP.  So\r\n" );
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document.write( "b²-a² = c²-b²\r\n" );
document.write( "b²+b² = a²+c²\r\n" );
document.write( "2b² = a²+c²\r\n" );
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document.write( "Add 2ab+2ac+2bc to both sides:\r\n" );
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document.write( "2ab+2b²+2ac+2bc = 2ac+2bc+a²+2ab+c²\r\n" );
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document.write( "Write 2ac as ac+ac, 2bc as bc+bc, 2ab as ab+ab\r\n" );
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document.write( "2ab+2b²+2ac+2bc = ac+ac+bc+bc+a²+ab+ab+c²\r\n" );
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document.write( "Rearrange the terms on the right:\r\n" );
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document.write( "2ab+2b²+2ac+2bc = ac+bc+a²+ab+bc+c²+ab+ac\r\n" );
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document.write( "Rewrite some of the terms by changing the order\r\n" );
document.write( "of factors:\r\n" );
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document.write( "2ba+2b²+2ca+2cb = ca+cb+a²+ab+cb+c²+ab+ac\r\n" );
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document.write( "Factor 2 out of the left side and group the terms on the right:\r\n" );
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document.write( "2(ba+b²+ca+cb) = (ca+cb+a²+ab) + (cb+c²+ab+ac)\r\n" );
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document.write( "Factor each expression in parentheses by grouping:\r\n" );
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document.write( "2[(ba+b²)+(ca+cb)] = [(ca+cb)+(a²+ab)] + [(cb+c²)+(ab+ac)]\r\n" );
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document.write( "2[b(a+b)+c(a+b)] = [c(a+b)+a(a+b)] + [c(b+c)+a(b+c)]\r\n" );
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document.write( "2(b+c)(a+b) = (c+a)(a+b) + (c+a)(b+c)\r\n" );
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document.write( "Divide through by (c+a)(b+c)(a+b) assuming that it is not 0.\r\n" );
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document.write( "Write  as \r\n" );
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document.write( "Rearranging the terms:\r\n" );
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document.write( "Therefore\r\n" );
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document.write( ", ,  are in AP, because\r\n" );
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document.write( "three terms are in AP, if and only if\r\n" );
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document.write( "  \r\n" );
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document.write( "Edwin
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