The sum of n terms of an arithmetic series is 5n^2-11n for all values of n. Detemine the common difference.

To determine the common difference, we'll have to consider
to consecutive terms of the a.p.

We'll
get:

an - an-1 = d

We know,
from enunciation, that the sum of n terms of the a.p.
is:

a1 + a2 + ... + an = 5n^2 -
11n

We'll determine an by subtracting both sides the sum:
a1 + a2 + .... + an-1:

an = 5n^2 - 11n - (a1 + a2 + .... +
an-1)

But a1 + a2 + .... + an-1 = 5(n-1)^2 -
11(n-1)

an = 5n^2 - 11n - 5(n-1)^2 +
11(n-1)

We'll expand the
squares:

an = 5n^2 - 11n - 5n^2 + 10n - 5 + 11n -
11

We'll combine and eliminate like
terms:

an  = 10n - 16

Knowing
the general term an, we can determine any term of the arithmetical
series.

a1 = 10*1 - 16

a1 = 10
- 16

a1 = -6

a2 = 10*2 -
16

a2 = 20 - 16

a2 =
4

a3 = 10*3 - 16

a3 =
14

The common difference is the difference between 2
consecutive terms:

a2 - a1 = 4 + 6 =
10

d = 10

We can verify and
we'll get a3 = a2 + d

14 = 4 +
10

14 = 14

So,
the common difference of the given arithmetic series is d =
10.