document.write( "Question 3840: the numbers a, b, c are the terms of a geometric sequence so that
\n" ); document.write( "a+b+c=35 and abc=1000\r
\n" ); document.write( "\n" ); document.write( "find a, b and c
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Algebra.Com's Answer #1679 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
a nice little question :-)\r
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\n" ); document.write( "\n" ); document.write( "Let first term = a
\n" ); document.write( "then second term = ar
\n" ); document.write( "and third term = \"ar%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "where r is the common ratio.\r
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\n" ); document.write( "\n" ); document.write( "And we know that all 3 terms multiplied together = 1000. So,\r
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\n" ); document.write( "\n" ); document.write( "\"a%28ar%29%28ar%5E2%29+=+1000\"
\n" ); document.write( "\"a%5E3r%5E3+=+1000\"
\n" ); document.write( "\"%28ar%29%5E3+1000\"\r
\n" ); document.write( "\n" ); document.write( "--> ar = cube root of 1000
\n" ); document.write( "ie ar = 10.\r
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\n" ); document.write( "\n" ); document.write( "And ar, by definition is the number b. So we have one found.\r
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\n" ); document.write( "\n" ); document.write( "Since abc = 1000 and b=10, then ac=100. We shall use this in a moment.\r
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\n" ); document.write( "\n" ); document.write( "Also, a+b+c=35. So now this becomes a+c=25 or a=25-c. Now sub this into the ac=100, to give (25-c)c=100.\r
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\n" ); document.write( "\n" ); document.write( "Multiplying this out, we get a quadratic, \"c%5E2-25c%2B100=0\" which factorises to (c-5)(c-20)=0, so c=5 or c=20. This means that a=20 or a=5...ie the reverse solution. So,\r
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\n" ); document.write( "\n" ); document.write( "a=5
\n" ); document.write( "b=10
\n" ); document.write( "c=20\r
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\n" ); document.write( "\n" ); document.write( "Common ratio is 2.
\n" ); document.write( "and Adding them --> 35
\n" ); document.write( "multiplying them --> 1000.\r
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\n" ); document.write( "\n" ); document.write( "jon
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