Chapter 8.2, Problem 33E

To determine

To find:

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

Answer to Problem 33E

Thefirst five terms are $22.45$ , $20.725$ , $19$ , $17.275$ , $15.55$ .

Explanation of Solution

Given information:

  $a_3=19$ and $a_{15}=-1.7$

The nth term of the AP series is:

  $a_n=a_1+\left(n-1\right)d$

Substitute $n=3$ in the above formula:

  $\begin{array}{c}{a}_3=a_1+\left(3-1\right)d\\ 19=a_1+2d\mathrm{............}\left(1\right)\end{array}$

Similarly,for $n=15$ .

  $\begin{array}{c}{a}_{15}=a_1+\left(15-1\right)d\\ -1.7=a_1+14d\mathrm{............}\left(2\right)\end{array}$

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

  $\begin{array}{c}{a}_1+14d-\left(a_1+2d\right)=-1.7-19\\ 14d-2d=-20.7\\ 12d=-20.7\\ d=-1.725\end{array}$

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

  $\begin{array}{c}19=a_1+2\left(-1.725\right)\\ -38=a_1-3.45\\ a_1=19+3.45\\ a_1=22.45\end{array}$

To find the first five term add $-1.725$ in each term.

  $\begin{array}{c}{a}_2=a_1+d\\ =22.45-1.725\\ =20.725\end{array}$

  $\begin{array}{c}{a}_3=a_2+d\\ =20.725-1.725\\ =19\end{array}$

  $\begin{array}{c}{a}_4=a_3+d\\ =19-1.725\\ =17.275\end{array}$

  $\begin{array}{c}{a}_5=a_4+d\\ =17.275-1.725\\ =15.55\end{array}$

Hence, the first five terms are $22.45$ , $20.725$ , $19$ , $17.275$ , $15.55$ .

Arithmetic sequence is a linear function.

$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$22.45$	$20.725$	$19$	$17.275$	$15.55$

To draw a graph with the above information: