a1=3xy^ 5
an= (2x/ y )a(n-1)
then
a2=(2x/ y )a(1)=
(2x/ y )* 3xy^ 5=6(x^ 2)(y^ 4)
a2=6(x^ 2)(y^ 4)
a3=(2x/ y )a(2)=
(2x/ y )* 6(x^ 2)(y^ 4)=12(x^ 3)(y^ 3)
a3=12(x^ 3)(y^ 3)
a4=(2x/ y )a(3)=(2x/
y )*12(x^ 3)(y^ 3)=24(x^ 4)(y^ 2)
The answer is
a4=24(x^ 4)(y^ 2)
