The sum of the given arithmetic series 5+6+7+ . . . . . . +11 is 56.
Suppose we are given an arithmetic series:
a(1) + a(2) + a(3) + a(4) + .... + a(n)
Then, each term of the series follow the general formula:
a(n) = a(1) + (n-1) . d
Where:
a(n) = nth term
a(1) = first term
d = common difference = a(n) - a(n-1)
We are given:
5+6+7+ . . . . . . +11
Parameters we get from this series:
a(1) = 5
d = 6 - 5 = 1
a(n) = 11
We need to find n that results in a(n) = 11
Substitute a(n) = 11, a(1) = 5, and d = 1:
a(n) = a(1) + (n-1) . d
11 = 5 + (n-1) . 1
11 = n + 4
n = 7
The sum of the first n terms of an arithmetic series is:
S(n) = n/2 [ a(1) + a(n) ]
S(7) = 7/2 (5 + 11)
= 7 . 8 = 56
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