The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

$$\small{\text{We have $t_x=t_5=9$ and $t_y=t_{32}=-84$ and we want $t_z=t_{23}=?$
}}\\\\
\small{\text{$\boxed{~~t_z = t_x\cdot \left( \dfrac{y-z}{y-x}\right)
+t_y\cdot \left(\dfrac{z-x}{y-x}\right) ~~}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
t_{23} &=& t_5\cdot \left( \dfrac{32-23}{32-5}\right)
+t_{32}\cdot \left(\dfrac{23-5}{32-5}\right) \\\\
t_{23} &=& 9\cdot \left( \dfrac{9}{27}\right)
-84\cdot \left(\dfrac{18}{27}\right) \\\\
t_{23} &=& 9\cdot \left( \dfrac{1}{3}\right)
-84\cdot \left(\dfrac{2}{3}\right) \\\\
t_{23} &=& -\dfrac{159}{3}\\\\
\mathbf{t_{23}} & \mathbf{=} & \mathbf{-53}\\
\hline
\\
\end{array}
$}}\\\\$$

The 23rd term is -53

We have the following system

9 = a₁ + d(5 -1)

-84 = a₁ + d(32 -1)       simplifying, we have

9 = a₁ + 4d

-84 = a₁ + 31d    subtract the second equation from the first

93 = -27d     divide both sides by -27

d = -31/9

Using the first equation to find a₁, we have

9  = a₁ + 4(131/9) 

a₁ = 9 - 4(-31/9) = 205/9    and this is the first term

So....the 23rd term is given by

205/9 + (-31/9)(23-1) = -53