document.write( "Question 974872: The value of xyz is 15/2 if a, x, y, z, b in AP.
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Algebra.Com's Answer #596729 by Edwin McCravy(20056) $\"\"$ $\"About$

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The value of xyz is 15/2 if a, x, y, z, b in AP.
\n" ); document.write( "While xyz is 18/5 if a, x, y, z, b in HP.
\n" ); document.write( "If a, b are +ve integers then find them.
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document.write( "Let d be the common difference, then \r\n" );
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document.write( "a=y-2d, x=y-d, y, z=y+d, b=y+2d\r\n" );
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document.write( "xyz = (y-d)y(y+d) = 15/2\r\n" );
document.write( "         y(y²-d²) = 15/2\r\n" );
document.write( "           y³-d²y = 15/2\r\n" );
document.write( "         2y³-2d²y = 15\r\n" );
document.write( "      2y³-2d²y-15 = 0\r\n" );
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document.write( "Since a and b are positive integers, a+b is a positive integer,\r\n" );
document.write( "therefore a+b = y-2d+y+2d = 2y is a positive integer.\r\n" );
document.write( "Let 2y = p, a positive integer.  Substitute y = p/2\r\n" );
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document.write( "2(p/2)³-2d²(p/2)-15 = 0\r\n" );
document.write( "       2p³/8-d²p-15 = 0\r\n" );
document.write( "        p³/4-d²p-15 = 0\r\n" );
document.write( "         p³-4d²p-60 = 0    \r\n" );
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document.write( "Since p³ = 4d²p+60, p³ is a multiple of 4,\r\n" );
document.write( "so is p.  4 is the only multiple of 4 which is a factor\r\n" );
document.write( "of 60 and thus that can be a rational solutional to the cubic.  \r\n" );
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document.write( "We use synthetic division with p=4:\r\n" );
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document.write( "4 | 1  0    -4d²      -60\r\n" );
document.write( "  |    4      16  64-16d²\r\n" );
document.write( "    1  4  16-4d²   4-16d²\r\n" );
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document.write( "That is a solution if and only if 4-16d² = 0\r\n" );
document.write( "                                   -16d² = -4\r\n" );
document.write( "                                      d² = 1/4\r\n" );
document.write( "                                       d = ±1/2\r\n" );
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document.write( "So we have p = 4, d = ±1/2\r\n" );
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document.write( "So y = p/2 = 4/2 = 2\r\n" );
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document.write( "a=y-2d, x=y-d, y, z=y+d, b=y+2d\r\n" );
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document.write( "Using d = 1/2\r\n" );
document.write( "a=2-2(1/2)=1, x=2-(1/2)=3/2, y=2, z=2+(1/2)=5/2, b=3\r\n" );
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document.write( "So the positive integers are a=1 and b=3\r\n" );
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document.write( "Using d = -1/2\r\n" );
document.write( "a=2-2(-1/2)=3, x=2-(-1/2)=5/2, y=2, z=2+(-1/2)=3/2, b=1\r\n" );
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document.write( "So the positive integers are a=3 and b=1\r\n" );
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document.write( "Either way the integers are 1 and 3.\r\n" );
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document.write( "Edwin

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