Question 1070612
There is a typo there.
{{{S[n]=5n^2-11}}} cannot be the sum of n terms of an arithmetic series,
but {{{S[n]=5n^2-11n}}} could be .
You could use either expression to find the common difference {{{d=10}}} ,
but only a sum of the form {{{an^2+bn}}} could work to find an actual
arithmetic series, with a first term {{{a[1]}}} independent of {{{n}}} .
 
A WAY TO SOLVE IT WITH {{{S[n]=5n^2-11n}}} :
The sum of {{{n-1}}} terms is
{{{S[n-1]=5(n-1)^2-11(n-1)}}} 
{{{S[n-1]=5(n^2-2n+1)-11n+11}}} 
{{{S[n-1]=5n^2-10n+5-11n+11}}} 
{{{S[n-1]=5n^2-21n+16}}} 
and since {{{S[n]=S[n-1]+a[n]}}} where {{{a[n]}}} is term number {{{n}}} ,
we can find an expression for {{{a[n]=S[n]-S[n-1]}}} .
{{{a[n]=5n^2-11n-(5n^2-21n+16)}}}
{{{a[n]=5n^2-11n-5n^2+21n-16)}}}
{{{a[n]=10n-16)}}}
{{{a[n]=10n-10-6)}}}
{{{a[n]=10(n-1)-6)}}}
That gives us {{{highlight(d=10)}}} as the common difference,
and {{{a[1]=-6}}} as the first term.
We know that in an arithmetic sequence
with a first term {{{a[1]}}} ,
and a common difference {{{d}}} ,
term number {{{n}}} is
{{{a[n]=a[1]+d(n-1)}}} ,
so {{{d}}} is the coefficient of {{{(n-1)}}} ,
and the other term is {{{a[1]}}} .
 
WHY WE KNOW {{{S[n]=5n^2-11}}} IS WRONG:
We have the "formula" {{{S[n]=(2a[1]+(n-1)d)n/2}}}
In that expression {{{n}}} is a factor.
The expression is a quadratic (degree 2) polynomial with {{{n]}} as a factor,
meaning no independent term.
{{{S[n]=(2a[1]+nd-d)n/2}}}
{{{S[n]=((2a[1]-d)+nd)n/2}}}
{{{S[n]=(dn^2+(2a[1]-d)n)/2}}}
{{{S[n]=(d/2)n^2+((2a[1]-d)/2)n}}}