Question 1091923
{{{a[n]=a[1]+(n-1)d}}}
So,
{{{a[3]=a[1]+(3-1)d}}}
1.{{{8=a[1]+2d}}}
and
{{{a[6]=a[1]+(6-1)d}}}
2.{{{2=a[1]+5d}}}
Subtract 1 from 2,
{{{a[1]+5d-a[1]-2d=2-8}}}
{{{3d=-6}}}
{{{d=-2}}}
Now use either equation to solve for {{{a[1]}}}.
Then solve for,
{{{a[30]=a[1]+(30-1)(-2)}}}
{{{a[30]=a[1]+29(-2)}}}
{{{a[30]=a[1]-58}}}