The sum of n terms of an arithmetic series is 5n^2-11n for all values of n. Detemine the common difference.
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
Therefore the common difference is 10.
To determine the common difference, we'll have to consider to consecutive terms of the a.p.
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.