document.write( "Question 1062664: Please could be found the rule of such a sequence:
\n" ); document.write( "72, 12, 6, 4, 3, 2, 2 ?
\n" ); document.write( "Please don't modify the given series.
\n" ); document.write( "The similar problem with 79, 24, 12... is much easier.
\n" ); document.write( "What might be the next number?
\n" ); document.write( "Thanks in advance,
\n" ); document.write( "Vlad \n" ); document.write( "
Algebra.Com's Answer #677631 by Edwin McCravy(20065)\"\" \"About 
You can put this solution on YOUR website!
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document.write( " The recursion formula is:\r\n" );
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document.write( " a(n) = a(n-3)/a(n-1) \r\n" );
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document.write( "72, 24, 12, 6, 4, 3, 2, 2, 3/2, 4/3, 3/2\r\n" );
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document.write( " a(4) =  a(1)/a(3) = 72÷12  = 6     \r\n" );
document.write( " a(5) =  a(2)/a(4) = 24÷6   = 4     \r\n" );
document.write( " a(6) =  a(3)/a(5) = 12÷4   = 3     \r\n" );
document.write( " a(7) =  a(4)/a(6) =  6÷3   = 2     \r\n" );
document.write( " a(8) =  a(5)/a(7) =  4÷2   = 2     \r\n" );
document.write( " a(9) =  a(6)/a(8) =  3÷2   = 3/2    \r\n" );
document.write( "a(10) =  a(7)/a(9) =  2÷3/2 = 2×(2/3) = 4/3   \r\n" );
document.write( "a(11) = a(8)/a(10) =  2÷4/3 = 2×(3/4) = 3/4 \r\n" );
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document.write( "The 9th term, a(9) = 3/2\r\n" );
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document.write( "Edwin
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