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
$$\small{\text{
If $b$ is positive, what is the value of $b$ in the geometric sequence $9, a , 4, b$?}}\\
\small{\text{
Express your answer as a common fraction.
}}$$
$$\small{\text{We have $ u_1 = 9,~ u_2 = a, ~ u_3 = 4, and ~ u_4 = b
$ ~~\boxed{ \dfrac{u_n}{u_{n-1}} = constant. }
}}\\\\
\small{\text{ $
\begin{array}{rcl}
\dfrac{u_2}{u_1} &=& \dfrac{u_3}{u_2} \\\\
\dfrac{a}{9} &=& \dfrac{4}{a} \\\\
a^2 &=& 4\cdot 9 = 36 \\\\
\mathbf{a} & \mathbf{=} & \mathbf{6}\\\\\\
\dfrac{u_3}{u_2} &=& \dfrac{u_4}{u_3} \\\\
\dfrac{4}{a} &=& \dfrac{b}{4} \\\\
\dfrac{4}{6} &=& \dfrac{b}{4} \\\\
b &=& \dfrac{16}{6} \\\\
\mathbf{b} & \mathbf{=} & \mathbf{\dfrac{8}{3}}
\end{array}
$}}\\\\$$