Chapter 8, Problem 11CR

To determine

Find the $n\text{th}$ term of the given sequence and graph the first six terms.

Answer to Problem 11CR

The $n\text{th}$ term $a_n=8-\left(n-1\right)4$

Explanation of Solution

Given:

The given data is $a_6=-12,a_{12}=-36$ .

Calculation:

The general rule for the $n\text{th}$ term is $a_n=a_1+\left(n-1\right)d$ .

Where $a_1\to$ first term, $d\to$ common difference, $n\to$ number of terms.

Substitute the value of $a_6=-12,\text{and}n=6$ in $n\text{th}$ term formula.

The $\text{6th}$ term

  $\begin{array}{l}{a}_6=a_1+\left(6-1\right)d\\ \Rightarrow -12=a_1+5d---\left(i\right)\end{array}$

Substitute the value of $a_{12}=-36,\text{and}n=12$ in $n\text{th}$ term formula.

The $\text{6th}$ term

  $\begin{array}{l}{a}_{12}=a_1+\left(12-1\right)d\\ \Rightarrow -36=a_1+11d---\left(ii\right)\end{array}$

Subtract $\left(ii\right)$ from $\left(i\right)$ .

  $\begin{array}{l}\underset{\_}{\begin{array}{l}-12=a_1+5d\\ -36=a_1+11d\\ +--\end{array}}\\ 24=0-6d\\ \Rightarrow -6d=24\\ \Rightarrow \frac{-6}{-6}d=\frac{24}{-6}\\ \Rightarrow d=-4\end{array}$

Substitute the value of $d=-4$ in equation $\left(i\right)$ .

  $\begin{array}{l}-12=a_1+5d\\ \Rightarrow -12=a_1+5\left(-4\right)\\ \Rightarrow -12=a_1-20\\ \Rightarrow -12+20=a_1-20+20\\ \Rightarrow 8=a_1\end{array}$

Now $n\text{th}$ term of the sequence is

  $\begin{array}{l}{a}_n=8+\left(n-1\right)\left(-4\right)\\ \Rightarrow a_n=8-\left(n-1\right)4\end{array}$

The first $6$ terms are

  $\begin{array}{l}{a}_1=8\\ a_2=a_1+d=8+\left(-4\right)=4\\ a_3=a_2+d=4+\left(-4\right)=0\\ a_4=a_3+d=0+\left(-4\right)=-4\\ a_5=a_4+d=-4+\left(-4\right)=-8\\ a_6=a_5+d=-8+\left(-4\right)=-12\end{array}$

Graph of the first six terms is given below.