Question 1186959

 

{{{a[1]=8}}}
{{{a[n]=34}}}
The sum of the first six terms is {{{58}}}.
  

Sum of the first six terms:

{{{(6/2)(2*8+5*d)=58}}}

{{{3(16+5d)=58}}}

{{{48+15d=58}}}

{{{15d=10}}}

{{{d=2/3}}}

plug it in last term formula and solve for {{{n}}}:

{{{34=8+(n-1)d}}}

{{{26=(n-1)(2/3)}}}

{{{39=n-1}}}

{{{ n=40}}}