Question 1097287
Let the consecutive numbers be {{{N-2}}}, {{{N}}}, {{{N+2}}}, and {{{N+4}}}.
So you know,
{{{N-2+N+N+2+N+4=36}}}
{{{4N+4=36}}}
{{{4N=32}}}
{{{N=8}}}
So then, the average would be,
{{{A=(N-2+N+4)/2=(2N+2)/2=N+1=9}}}