document.write( "Question 1209387: Let
\n" ); document.write( "a + ar + ar^2 + ar^3 + ...
\n" ); document.write( "be an infinite geometric series. The sum of the series is 3. The sum of the cubes of all the terms is 5. Find the common ratio.
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Algebra.Com's Answer #848731 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The given series is

\n" ); document.write( "a+ar+ar^2+ar^3+...

\n" ); document.write( "The sum of the series is 3:

\n" ); document.write( "[1] \"a%2F%281-r%29=3\"

\n" ); document.write( "The series consisting of the cubes of the terms of the given series is

\n" ); document.write( "a^3r^3+a^3r^6+a^3r^9+...

\n" ); document.write( "The sum of that series is 5:

\n" ); document.write( "[2] \"a%5E3%2F%281-r%5E3%29=5\"

\n" ); document.write( "To find the common ratio r, solve [1] for a in terms of r and substitute in [2].

\n" ); document.write( "\"a=3%281-r%29\"

\n" ); document.write( "\"%283%281-r%29%29%5E3%2F%281-r%5E3%29=5\"

\n" ); document.write( "\"27%281-3r%2B3r%5E2-r%5E3%29=5%281-r%5E3%29\"

\n" ); document.write( "\"27-81r%2B81r%5E2-27r%5E3=5-5r%5E3\"

\n" ); document.write( "\"22r%5E3-81r%5E2%2B81r-22=0\"

\n" ); document.write( "By inspection, r=1 is one solution to that equation. However r=1 produces an infinite geometric series that has no sum.

\n" ); document.write( "Use synthetic division to remove the root x=1 to find the other two roots.
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document.write( "  1 | 22 -81  81 -22\r\n" );
document.write( "    |     22 -59  22\r\n" );
document.write( "    +---------------\r\n" );
document.write( "      22 -59  22   0

\n" ); document.write( "The remaining quadratic is \"22x%5E2-59x%2B22\"

\n" ); document.write( "Use the quadratic formula to find that the other two solutions are

\n" ); document.write( "\"%2859%2B-sqrt%281545%29%29%2F44\"

\n" ); document.write( "There are two possible values for the common ratio r:

\n" ); document.write( "ANSWERS:
\n" ); document.write( "\"%281%2F44%29%2859%2Bsqrt%281545%29%29\"
\n" ); document.write( "\"%281%2F44%29%2859-sqrt%281545%29%29\"

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