Answer:
solution
given=3a+3a^2+3a^3+....+...+...3an
first term(t1)=3a
second term(t2)=3a^2
third term(t3)=3a^3
for arithmetic mean ,d1=d2
d1=t2_t1=3a^2_3a=3a(a_1)
d2=3a^3_3a^2=3a^2(a_1)
here d1 is not equal to d2.
so it's not an arithmetic series.
Again,for geometric mean,r1=r2
r1=d2/d1=3a^2/3a=a
r2=d3/d2=3a^3/3a^2=a
here ,r1 is equal to r2.
so ,it's an geometric series.
