arithmetic sequence

Let $a_n$ be the $n$th term of the sequence. ($a_n=a_1+n*(d-1)$ where $d$ is the common difference, $a_1$ is the first term, and $n$ is the number of the term you want to find out)

We are given that $a_5=9$ and $a_{32}=84$.

From those, we can get the equations $9=a_1+4d$ and $84=a_1+31d$.

(Math time!)

$a_1+31d = 84$

$a_1+4d   = 9$

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        $27d =  75$

            $d = \frac{25}{9}$

Since $9=a_1+4d$:

$9=a_1+\frac{100}{9}$

$a_1=\frac{-19}{9}$

Finally, $a_23=a_1+22*d$, so:

$a_{23}=\frac{-19}{9}+22*\frac{25}{9}$

$a_{23}=\frac{550-19}{9}$

$a_{23}=\frac{531}{9}$

$a_{23}=\boxed{59}$.

The fifth term of an arithmetic sequence is 9 and the 32nd term is 84.

What is the 23rd term?

\(\begin{array}{|lrcll|} \hline (1) & a_i &=& a_1+(i-1)d \\ (2) & a_j &=& a_1+(j-1)d \\ (3) & a_k &=& a_1+(k-1)d \\ \hline & a_1 = a_i-(i-1)d &=& a_j-(j-1)d \\ & a_i-(i-1)d &=& a_j-(j-1)d \\ & a_j-a_i &=& (j-1)d-(i-1)d \\ & a_j-a_i &=& \Big((j-1)-(i-1)\Big)d \\ (4) & \mathbf{a_j-a_i} &=& \mathbf{(j-i)d} \\ \hline & a_1 = a_j-(j-1)d &=& a_k-(k-1)d \\ & a_j-(j-1)d &=& a_k-(k-1)d \\ & a_k-a_j &=& (k-1)d-(j-1)d \\ & a_k-a_j &=& \Big((k-1)-(j-1)\Big)d \\ (5)& \mathbf{a_k-a_j} &=& \mathbf{(k-j)d} \\ \hline \dfrac{(4)}{(5)}: & \dfrac{a_j-a_i}{a_k-a_j}&=& \dfrac{(j-i)d}{(k-j)d} \\\\ & \dfrac{a_j-a_i}{a_k-a_j}&=& \dfrac{(j-i)}{(k-j)} \\\\ & \ldots \\ & \mathbf{a_k} &=& \mathbf{a_j*\dfrac{(k-i)}{(j-i)}+a_i*\dfrac{(k-j)}{(i-j)}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline i=5 \qquad a_i &=& 9 \\ j=32 \qquad a_j &=& 84 \\ k=23 \qquad a_k &=& \ ? \\ \hline \mathbf{a_k} &=& \mathbf{a_j*\dfrac{(k-i)}{(j-i)}+a_i*\dfrac{(k-j)}{(i-j)}} \\ a_{23} &=& 84*\dfrac{(23-5)}{(32-5)}+9*\dfrac{(23-32)}{(5-32)} \\ a_{23} &=& 84*\dfrac{18}{27}+9*\dfrac{-9}{-27} \\ a_{23} &=& \dfrac{84*18+9*9}{27} \\ a_{23} &=& \dfrac{84*2+9}{3} \\ a_{23} &=& \dfrac{177}{3} \\ \mathbf{a_{23}} &=& \mathbf{59} \\ \hline \end{array}\)