Chapter 8.2, Problem 32E

To determine

To find:

Thefirst five terms of the arithmetic sequence also verify the results using graphing utility.

Answer to Problem 32E

Thefirst five terms are $7$ , $0$ , $-7$ , $-14$ , $-22$ .

Explanation of Solution

Given information:

  $a_6=-38$ and $a_{11}=-73$

The nth term of the AP series is:

  $a_n=a_1+\left(n-1\right)d$

Substitute $n=6$ in the above formula:

  $\begin{array}{c}{a}_6=a_1+\left(6-1\right)d\\ -38=a_1+5d\mathrm{............}\left(1\right)\end{array}$

Similarly,for $n=11$ .

  $\begin{array}{c}{a}_{11}=a_1+\left(11-1\right)d\\ -73=a_1+10d\mathrm{............}\left(2\right)\end{array}$

On subtracting equation $\left(1\right)$ from equation $\left(2\right)$ .

  $\begin{array}{c}{a}_1+10d-\left(a_1+5d\right)=-73-\left(-38\right)\\ 10d-5d=-73+38\\ 5d=-35\\ d=-7\end{array}$

By substituting the value of $d$ in equation $\left(1\right)$ .

  $\begin{array}{c}-38=a_1+5\left(-7\right)\\ -38=a_1-35\\ a_1=-38+35\\ a_1=-3\end{array}$

To find the first five terms add $-7$ in each term.

  $\begin{array}{c}{a}_2=a_1+d\\ =-3-7\\ =-10\end{array}$

  $\begin{array}{c}{a}_3=a_2+d\\ =-10-7\\ =-17\end{array}$

  $\begin{array}{c}{a}_4=a_3+d\\ =-17-7\\ =-24\end{array}$

  $\begin{array}{c}{a}_5=a_4+d\\ =-24-7\\ =-31\end{array}$

Hence, the first five terms are $-3$ , $-10$ , $-17$ , $-24$ , $-31$ .

Arithmetic sequence is a linear function.

$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$-3$	$-10$	$-17$	$-24$	$-31$

To draw a graph with the above information: