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
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