The fourth term of the sequence $3, \dfrac {3}{2}, 1...$ is:
My Attempt: Checking For $A.P$
First term$(a)=3$
Second term$(t_2)=\dfrac {3}{2}$
Third term$(t_3)=1$
This will be AP if $$t_3-t_2=t_2-t_1$$ $$1-\dfrac {3}{2}=\dfrac {3}{2}-3$$ $$\dfrac {-1}{2}=\dfrac {-3}{2}$$
So, this is not AP. AND also not GP.