document.write( "Question 858287: two arithmetic series are such that their common difference are 9 and 3 respectively. If their first terms are 2 and 5 respectively, find the number of terms of each series that would give a common sum. \n" ); document.write( "
Algebra.Com's Answer #517152 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
two arithmetic series are such that their common difference are 9 and 3 respectively. If their first terms are 2 and 5 respectively, find the number of terms of each series that would give a common sum. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let Sn be the sum of first series d=9, a=2\r
\n" ); document.write( "\n" ); document.write( "S'n be the sum of second series. d=3, a=5\r
\n" ); document.write( "\n" ); document.write( "Sn = n/2(2a1+(n-1)d)
\n" ); document.write( "Sn = n/2 (2*2+(n-1) 9)\r
\n" ); document.write( "\n" ); document.write( "Sn = n/2( 4+9n-9)
\n" ); document.write( "Sn = n/2(4n-5)\r
\n" ); document.write( "\n" ); document.write( "Find S'n\r
\n" ); document.write( "\n" ); document.write( "S'n = n/2(2*5+(n-1)*3)\r
\n" ); document.write( "\n" ); document.write( "S'n = n/2(10+3n-3)\r
\n" ); document.write( "\n" ); document.write( "S'n = n/2(3n+7)\r
\n" ); document.write( "\n" ); document.write( "Sn=S'n\r
\n" ); document.write( "\n" ); document.write( "n/2(4n-5)=n/2(3n+7)\r
\n" ); document.write( "\n" ); document.write( "4n-5 = 3n+7\r
\n" ); document.write( "\n" ); document.write( "4n-3n=12
\n" ); document.write( "n=12\r
\n" ); document.write( "\n" ); document.write( "For the 12 th. term the sums are equal\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );