Chapter 8.2, Problem 30E

To determine

To calculate: The another term of the sequence

Answer to Problem 30E

Option A is correct.

Explanation of Solution

Given information:

  $a_{12}=43,d=6$ .

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

The $nth$ tem of an arithmetic sequence with first term $a_1$ and common difference $d$ is given by : $a_n=a_1+(n-1)d$ .

Calculation:

Step 1. Using general rule to find the first term .

  $\begin{array}{l}=a_n=a_1+(n-1)d\\ =a_{12}=a_1+\left(12-1\right)\left(6\right)\\ =43=a_1+\left[\left(12\times 6\right)-\left(1\times 6\right)\right]\\ =43=a_1+\left(72-6\right)\\ =43=a_1+\left(66\right)\\ =43-66=a_1\\ =-23=a_1\end{array}$

Step 2. Rule for $nth$ term.

  $\begin{array}{l}{a}_n=-23+\left(n-1\right)\left(6\right)\\ =-23+6n-6\\ =6n-29\end{array}$

Step 3. Using the rule formed in Step 2. will construct a table and then plot the points.

  $a_n=6n-29$

$n$	$1$	$2$	$3$	$4$	$5$	$6$
$a_n$	$\begin{array}{l}{a}_n=6(1)-29\\ =6-29\\ =-23\end{array}$	$\begin{array}{l}{a}_n=6(2)-29\\ =12-29\\ =-17\end{array}$	$\begin{array}{l}{a}_n=6(3)-29\\ =18-29\\ =-11\end{array}$	$\begin{array}{l}{a}_n=6(4)-29\\ =24-29\\ =-5\end{array}$	$\begin{array}{l}{a}_n=6(5)-29\\ =30-29\\ =1\end{array}$	$\begin{array}{l}{a}_n=6(6)-29\\ =36-29\\ =7\end{array}$

The another term of the sequence is $a_3=-11$ .