Question 1044435
{{{n+(n+1)+(n+2)=30}}}
{{{3n+3=30}}}
{{{n+1=10}}}
{{{n=9}}}


The first number of the sequence of numbers is 9 and common difference to reach each succeeding number is 1.


You are looking for {{{12+13+14+15+16}}} for the answer to the question.  The simple way to find this, for so few terms to sum, is {{{(5/2)(12+16)}}}
{{{(5/2)*28}}}
{{{5*14}}}
{{{highlight(70)}}}