Question 1085838
In a list of n consecutive odd numbers, the number 
of terms equals the first term. If six times the sum 
of the n numbers exceeds the square of their average 
by 2013 times the average, what is the value of the 
first term?
<pre>

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}
{{{Average = S[n]/n^""}}}</pre>
consecutive odd numbers, 
<pre>
{{{d = 2}}}
</pre>
the number of terms equals the first term.
<pre>
{{{n=a[1]}}}

Substituting those and simplifying

{{{S[n]=expr(n/2)(2n+(n-1)2)=expr(n/2)(2n+2n-2)=expr(n/2)(4n-2)=n(2n-1)}}}
{{{Average=S[n]/n^""=n(2n-1)/n=2n-1}}}
</pre>
If six times the sum of the n numbers exceeds the 
square of their average by 2013 times the average, 
<pre>
{{{6n(2n-1)=(2n-1)^2+2013(2n-1)}}}

Divide through by (2n-1), which cannot be 0,
since n cannot be 1/2.

{{{6n=(2n-1)+2013}}}

{{{6n=2n-1+2013}}}

{{{6n=2n+2012}}}

{{{4n=2012}}}

{{{n=503}}}

First term = {{{a[1]=n=503}}}

Edwin</pre>