Question 27288: Use the formula for an arithmetic series to find each sum.
40+42+44+...+68.
Thank you very much
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! 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
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