a1=7a0+3; a2=7a1+3=7(7a0+3)+3; a3=7a2+3=7(7(7a0+3)+3)+3, ...
So let's expand a3:
7(72a0+7×3+3)+3=73a0+72×3+7×3+3=73a0+3(1+7+72).
We can see a pattern: an=7na0+3(1+7+...+7n-1).
This incudes the GP=1+7+...+7n-1=(7n-1)/(7-1)=(7n-1)/6.
an=7na0+3(7n-1)/6=7na0+½(7n-1).
a0=2; a1=7a0+3=17; a2=7a1+3=122; a3=7a2+3=857.
Now test a3 using an=7na0+½(7n-1)=343×2+342/2=686+171=857. So the formula works.