document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=461296>Question 461296</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if a,b,c are in h.p then prove that bc/(b+c),ca/(c+a),ab/(a+b) are in h.p </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #316730 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=316730\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I'm guessing \"h.p.\" means harmonic progression? I recommend you do not use unusual abbreviations.\r<BR>\n" );
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document.write( "If a, b, c is a harmonic progression, then we can say that a = 1/k, b = 1/(k+d) and c = 1/(k+2d). We can replace the expressions for a, b, c into the next three terms:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{b+c}{bc} = \frac{ \frac{1}{k+d} \frac{1}{k+2d} }{\frac{1}{k+d} + \frac{1}{k+2d}} = \frac{1}{2k + 3d}\">\r<BR>\n" );
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document.write( "Similarly,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{c+a}{ca} = \frac{ \frac{1}{k+2d} \frac{1}{k} }{\frac{1}{k+2d} + \frac{1}{k}} = \frac{1}{2k + 2d}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{a+b}{ab} = \frac{ \frac{1}{k} \frac{1}{k+d} }{\frac{1}{k} + \frac{1}{k+d}} = \frac{1}{2k + d}\">\r<BR>\n" );
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document.write( "We see the third, second, and first terms (in that order) follow a harmonic progression because the reciprocals of each make an arithmetic progression with common difference d.  </SPAN>\n" );
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