document.write( "Question 703202: Pleas help me answer this problem : In an arithmetic progression.The sum of the first three terms is 18,and the sum of the squares of these terms is 126,Find the terms. \n" ); document.write( "

Algebra.Com's Answer #433354 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"d\"$ = common difference
\n" ); document.write( " $\"x\"$ = second term (I think that may make it easier)
\n" ); document.write( "first term = $\"x-d\"$
\n" ); document.write( "third term = $\"x%2Bd\"$
\n" ); document.write( "The sum of the first three terms is
\n" ); document.write( " $\"%28x-d%29%2Bx%2B%28x%2Bd%29=18\"$ --> $\"3x=18\"$ --> $\"highlight%28x=6%29\"$
\n" ); document.write( "The sum of the squares of these terms is
\n" ); document.write( " $\"%286-d%29%5E2%2B6%5E2%2B%286%2Bd%29%5E2=126\"$
\n" ); document.write( " $\"6%5E2-12d%2Bd%5E2%2B6%5E2%2B6%5E2%2B12d%2Bd%5E2=126\"$
\n" ); document.write( " $\"6%5E2%2B6%5E2%2B6%5E2%2B2d%5E2=126\"$
\n" ); document.write( " $\"3%2A6%5E2%2B2d%5E2=126\"$
\n" ); document.write( " $\"3%2A36%2B2d%5E2=126\"$
\n" ); document.write( " $\"3%2A36%2B2d%5E2=126\"$
\n" ); document.write( " $\"108%2B2d%5E2=126\"$ --> $\"2d%5E2=126-108\"$ --> $\"2d%5E2=18\"$ --> $\"d%5E2=18%2F2\"$ --> $\"d%5E2=9\"$
\n" ); document.write( "There are two solutions:
\n" ); document.write( "either $\"d=3\"$ or $\"d=-3\"$
\n" ); document.write( "Either way, the terms are
\n" ); document.write( " $\"6-3=highlight%283%29\"$ , $\"highlight%286%29\"$ and $\"6%2B3=highlight%289%29\"$ \n" ); document.write( "

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