Question 1126253

{{{a[n]=a[1]+(n-1)d}}}....using nth term formula, we have

{{{a[n]=a[1]+dn-d}}}

{{{a[n]=dn+(a[1]-d)}}}

since you are given

{{{a[n]=6n-17}}} , you see that

{{{(a[1]-d)=-17 }}} and {{{dn=6n}}}

=>{{{d=6}}}, is common difference

and

{{{a[1]-6=-17}}}
{{{a[1]=-17+6}}}
{{{a[1]=-11}}} is  its first term