Question 27288
a, a+d, a+2d, a+3d, ........, a+(k-1)d,....
is called the Standard Arithmetic progression.
And a + (a+d) + (a+2d) + (a+3d) ........ + [a+(k-1)d]+ ....
is called the Standard Arithmetic Series.
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.
The formula for the sum to n terms of an arithmetic series is
Sn = [n(T1 + Tn)]/2
In the sum: 40+42+44+....+68
we have T1 = a = 40, the common difference, d = 2 and the nth term Tn = 68
To find n we apply the formula for the nth term which is
 Tn = [a + (n - 1)d]
Therefore  [a + (n - 1)d] = 68
            40 + (n - 1)x2 = 68
Dividing by 2 through out
            20+(n-1) = 34
              (n-1)  = 34 - 20(transposing, change side then change sign)
               n-1 = 14
                n = 14 + 1 = 15
Putting n = 15, T1 = 40, Tn = 68 in the formula
       Sn = [n(T1 + Tn)]/2 = 15X(40+68)/2 = 15x108/2 = 15x54 = 810

The required sum is 810