Chapter 8.2, Problem 12E

To determine

To compare: The terms of an arithmetic sequence when $d>0\text{ to when }d<0.$

Explanation of Solution

Given information:

The first term is $-3$ and each term is 6 less than the previous term.

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:

According to the question ,

  $d=6>0$

Let , the $nth$ term is

  $\begin{array}{l}{a}_n=a_1+(n-1)6\\ a_1=1\end{array}$

Now,

  $\begin{array}{l}{a}_n=1+(n-1)6\\ =1+6n-6\\ =6n-5\end{array}$

The arithmetic sequence will be as follows:-

  $\begin{array}{l}{a}_2=6(2)-5\\ =12-5\\ =7\\ a_3=6(3)-5\\ =18-5\\ =13\end{array}$

Therefore , we get increasing sequence is $1,7,13,\mathrm{.....}$

Now, if

  $d=-6<0$

Let , the $nth$ term is

  $\begin{array}{l}{a}_n=a_1+(n-1)(-6)\\ a_1=1\end{array}$

Now,

  $\begin{array}{l}{a}_n=1+(n-1)(-6)\\ =1-6n+6\\ =-6n+7\end{array}$

The arithmetic sequence will be as follows:-

  $\begin{array}{l}{a}_2=-6(2)+7\\ =-12+7\\ =-5\\ a_3=-6(3)+7\\ =-18+7\\ =-11\end{array}$

Therefore , we get decreasing sequence is $1,-5,-11,\mathrm{.....}$