Chapter 9.2, Problem 14E

To determine

To find the first five terms of the sequence $a_n=100-3n$ , where $n\ge 1$ and TO-DETERMINE whether the sequence is arithmetic, if yes, then find its common difference.

Answer to Problem 14E

  $a_1=97,a_2=94,a_3=91,a_4=88,a_5=85$

The sequence $a_n=100-3n$ is arithmetic and the common difference $d=-3$ .

Explanation of Solution

Given information :

Let the sequence be $a_n=100-3n$ .

Concept used:

In a sequence $a_1,a_2,a_3,a_4,\dots \dots ,a_n,\dots$ if the difference between the consecutive terms are the same then it is defined as Arithmetic Sequence.

Here, $a_2-a_1=a_3-a_2=a_4-a_3=\dots \dots =d.$ The number $d$ is the common difference of the arithmetic sequence.

Calculations:

The first five terms of the sequence $a_n=100-3n$ are:

  $\begin{array}{l}{a}_1=100-3\left(1\right)=97\\ a_2=100-3\left(2\right)=100-6=94\\ a_3=100-3\left(3\right)=100-9=91\\ a_4=100-3\left(4\right)=100-12=88\\ a_5=100-3\left(5\right)=100-15=85\end{array}$

The terms of the sequence are $a_1=97,a_2=94,a_3=91,a_4=88,a_5=85,\dots ,$ .

In order TO-DETERMINE whether the sequence is arithmetic or not, calculate the difference of the consecutive terms:

  $\begin{array}{l}{a}_2-a_1=94-97=-3\\ a_3-a_2=91-94=-3\\ a_4-a_3=88-91=-3\\ a_5-a_4=85-88=-3\\ ⋮\end{array}$

And so on.

Since, $a_2-a_1=a_3-a_2=a_4-a_3=a_5-a_4=\dots \dots =-3=d$ , therefore the sequence is arithmetic and the common difference $d=-3$ .