document.write( "Question 858284: Three consecutive numbers of a G.p are such that there sum is 26 and there product is 216.Find the numbers \n" ); document.write( "

Algebra.Com's Answer #517155 by KMST(5328) $\"\"$ $\"About$

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THE FAST WAY:
\n" ); document.write( "If the product is $\"216\"$ ,
\n" ); document.write( " $\"216=6%5E3=%282%2A3%29%5E3=2%5E3%2A3%5E2\"$
\n" ); document.write( "The product of the three terms could be
\n" ); document.write( " $\"2%2A%282%2A3%29%2A%282%2A3%5E2%29=216\"$ with the terms being 2, 6, and 18 (or 18, 6, and 2)
\n" ); document.write( "or
\n" ); document.write( "the product could be
\n" ); document.write( " $\"3%2A%283%2A2%29%2A%283%2A2%5E2%29=216\"$ with the terms being 3, 6, and 12 (or 12, 6, and 3)
\n" ); document.write( "
\n" ); document.write( " $\"18%2B2%2B6=26\"$ while $\"12%2B6%2B3=21\"$ , so the numbers are
\n" ); document.write( " $\"highlight%282%29\"$ , $\"highlight%286%29\"$ , and $\"highlight%2818%29\"$ .
\n" ); document.write( "That gives as a solution quickly, without proving that it is the only solution.
\n" ); document.write( "
\n" ); document.write( "THE EXPECTED WAY (with lots of formulas and calculations):
\n" ); document.write( "We will call the first term we are looking for $\"b%5B1%5D=b\"$ ,
\n" ); document.write( "and the common ratio $\"r\"$ .
\n" ); document.write( "The formula for term number $\"n\"$ is $\"b%5Bn%5D=b%5B1%5D%2Ar%5E%28n-1%29\"$ .
\n" ); document.write( "The second and third terms would be
\n" ); document.write( " $\"b%5B2%5D=b%2Ar\"$ and $\"b%5B3%5D=b%2Ar%5E2\"$ .
\n" ); document.write( "Their product would be $\"b%2A%28b%2Ar%29%2A%28b%2Ar%5E2%29=b%5E3%2Ar%2A3=%28br%29%5E3\"$
\n" ); document.write( "We could calculate their sum as
\n" ); document.write( " $\"SUM=B%2Bbr%2Bbr%5E3=b%28r%5E2%2Br%2B1%29\"$
\n" ); document.write( "Otherwise, the formula for the sum of the first $\"n\"$ terms is
\n" ); document.write( " $\"SUM%5Bn%5D=b%5B1%5D%28r%5En-1%29%2F%28r-1%29\"$ .
\n" ); document.write( "The sum of the three terms would be
\n" ); document.write( " $\"SUM=b%28r%5E3-1%29%2F%28r-1%29=%28br%5E3-b%29%2F%28r-1%29\"$
\n" ); document.write( "Our equations are
\n" ); document.write( " $\"b%5E3%2Ar%2A3=216\"$ or $\"%28br%29%5E3=216\"$ and
\n" ); document.write( " $\"b%28r%5E3-1%29%2F%28r-1%29=26%29\"$ or $\"%28br%5E3-b%29%2F%28r-1%29=26%29\"$ or $\"b%28r%5E2%2Br%2B1%29=26\"$ .
\n" ); document.write( "From $\"%28br%29%5E3=216\"$ we get $\"br=root%283%2C210%29\"$ --> $\"br=6\"$ --> $\"system%28b=6%2Fr%2C%22or%22%2Cr=6%2Fb%29\"$
\n" ); document.write( "If we substitute either one into $\"b%28r%5E2%2Br%2B1%29=26\"$ we get an equation in one variable that we can solve.
\n" ); document.write( " $\"%286%2Fr%29%28r%5E2%2Br%2B1%29=26\"$ --> $\"6%28r%5E2%2Br%2B1%29=26r\"$ --> $\"6r%5E2%2B6r%2B6=26r\"$ --> $\"6r%5E2%2B6r%2B6-26r=0\"$ --> $\"6r%5E2-20r%2B6=0\"$ <--> $\"3r%5E2-10r%2B3=0\"$
\n" ); document.write( "No matter how we solve $\"6r%5E2-20r%2B6=0\"$ or $\"3r%5E2-10r%2B3=0\"$
\n" ); document.write( "we find $\"system%28r=3%2C%22or%22%2Cr=1%2F3%29\"$ .
\n" ); document.write( " $\"system%28r=3%2Cb=6%2Fr%29\"$ --> $\"system%28r=3%2Cb=6%2F3%29\"$ --> $\"highlight%28system%28r=3%2Cb=2%29%29\"$ --> the terms , in order, are 2, 6, and 18.
\n" ); document.write( " $\"system%28r=1%2F3%2Cb=6%2Fr%29\"$ --> $\"system%28r=3%2Cb=6%2F%281%2F3%29%29\"$ --> $\"system%28r=1%2F3%2Cb=6%2A3%29\"$ --> $\"highlight%28system%28r=1%2F3%2Cb=18%29%29\"$ --> the terms , in order, are 18, 6, and 2.
\n" ); document.write( "The terms (regardles of order are $\"highlight%282%29\"$ , $\"highlight%286%29\"$ , and $\"highlight%2818%29\"$ ,
\n" ); document.write( "and that is the only solution \n" ); document.write( "

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