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

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We are given that the sum of n terms of an arithmetic series is 5n^2-11n.

Now in an arithmetic series the difference between consecutive terms is the same.

We first find the sum of n+1 terms

This is given by S(n+1) =

5*(n+1)^2 - 11*(n+1)

=> 5*(n^2 + 2n +1) - 11n - 11

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

=> 5n^2 - n - 6

S(n + 1) - Sn = Tn

=> Tn = 5n^2 - n - 6 - 5n^2 + 11n

=> Tn = 10n - 6

Now T(n-1) = 10*(n-1) - 6 = 10n - 10 - 6

Tn - T(n-1) = 10n - 6 - 10n + 10 + 6

= 10

**Therefore the common difference is 10.**

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.**