document.write( "Question 27288: Use the formula for an arithmetic series to find each sum.\r
\n" ); document.write( "\n" ); document.write( "40+42+44+...+68.\r
\n" ); document.write( "\n" ); document.write( "Thank you very much \n" ); document.write( "
Algebra.Com's Answer #14819 by sdmmadam@yahoo.com(530)\"\" \"About 
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a, a+d, a+2d, a+3d, ........, a+(k-1)d,....
\n" ); document.write( "is called the Standard Arithmetic progression.
\n" ); document.write( "And a + (a+d) + (a+2d) + (a+3d) ........ + [a+(k-1)d]+ ....
\n" ); document.write( "is called the Standard Arithmetic Series.
\n" ); document.write( "Here T1 = a is the first term and Tn = [a + (n - 1)d] is the nth term of the Series and d is called the common difference.
\n" ); document.write( "The formula for the sum to n terms of an arithmetic series is
\n" ); document.write( "Sn = [n(T1 + Tn)]/2
\n" ); document.write( "In the sum: 40+42+44+....+68
\n" ); document.write( "we have T1 = a = 40, the common difference, d = 2 and the nth term Tn = 68
\n" ); document.write( "To find n we apply the formula for the nth term which is
\n" ); document.write( " Tn = [a + (n - 1)d]
\n" ); document.write( "Therefore [a + (n - 1)d] = 68
\n" ); document.write( " 40 + (n - 1)x2 = 68
\n" ); document.write( "Dividing by 2 through out
\n" ); document.write( " 20+(n-1) = 34
\n" ); document.write( " (n-1) = 34 - 20(transposing, change side then change sign)
\n" ); document.write( " n-1 = 14
\n" ); document.write( " n = 14 + 1 = 15
\n" ); document.write( "Putting n = 15, T1 = 40, Tn = 68 in the formula
\n" ); document.write( " Sn = [n(T1 + Tn)]/2 = 15X(40+68)/2 = 15x108/2 = 15x54 = 810\r
\n" ); document.write( "\n" ); document.write( "The required sum is 810\r
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