Question 1039414
Look at the sequence
{{{a}}}
{{{a-2}}}
{{{a-4}}
{{{a-6}}}
{{{a-8}}}
5 terms starting with a, each successive term removes 2. 
The sum would be,
{{{S=a+a-2+a-4+a-6+a-8}}}
{{{S=5a-20}}}
{{{S=5(a-4)}}}
 or in terms of the number of values,
{{{S=5(a-4)=n(a-(n-1))=n(a+1-n)}}}
So in your sequence,
{{{3/2+1/2-1/2-3/2-5/2-7/2-9/2-11/2=(1/2)(3+1-1-3-5-7-9-11)}}}
The starting term is {{{a=3}}} and {{{n=8}}},
so then,
{{{S=(1/2)8(3+1-8)}}}
{{{S=4(4-8)}}}
{{{S=4(-4)}}}
{{{S=-16}}}