If $b$ is positive, what is the value of $b$ in the geometric sequence $9, a , 4, b$? Express your answer as a common fraction.

$$\small{\text{ If $b$ is positive, what is the value of $b$ in the geometric sequence $9, a , 4, b$?}}\\ \small{\text{ Express your answer as a common fraction. }}$$

$$\small{\text{We have $ u_1 = 9,~ u_2 = a, ~ u_3 = 4, and ~ u_4 = b $ ~~\boxed{ \dfrac{u_n}{u_{n-1}} = constant. } }}\\\\ \small{\text{ $ \begin{array}{rcl} \dfrac{u_2}{u_1} &=& \dfrac{u_3}{u_2} \\\\ \dfrac{a}{9} &=& \dfrac{4}{a} \\\\ a^2 &=& 4\cdot 9 = 36 \\\\ \mathbf{a} & \mathbf{=} & \mathbf{6}\\\\\\ \dfrac{u_3}{u_2} &=& \dfrac{u_4}{u_3} \\\\ \dfrac{4}{a} &=& \dfrac{b}{4} \\\\ \dfrac{4}{6} &=& \dfrac{b}{4} \\\\ b &=& \dfrac{16}{6} \\\\ \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{8}{3}} \end{array} $}}\\\\$$

9 is the 1st term and 4 is the third

And the nth term is given by....    a₁(r)^(n-1 )    where a₁ is the first term....and r is the geometric multiplier.....so, we have

9(r)^(3- 1)  = 9(r)^(2)  =  4      divide both sides by  9

(r)^2 =  4/9    take the square root of both sides

r  = 2/3

So, b =  9(2/3)^(4 -1)  = 9(2/3)^3 =  8/3