document.write( "Question 74069: Find the value of x where 201, x, 1089 are three consecutive terms in a geometric sequence. \n" ); document.write( "
Algebra.Com's Answer #53071 by checkley75(3666)\"\" \"About 
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(201x)x=1089
\n" ); document.write( "201x^2-1089=0
\n" ); document.write( "using the quadratic equation
\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "we get
\n" ); document.write( "x=(0+-sqrt[0^2-4*201*-1089])/2*201
\n" ); document.write( "x=(+-sqrt875556)/402
\n" ); document.write( "x=(+-935.71)/402
\n" ); document.write( "x=935.71/402
\n" ); document.write( "x=2.32764 answer
\n" ); document.write( "proof
\n" ); document.write( "201*2.32764=468
\n" ); document.write( "468*2.32764=1089
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