Chapter 4.6, Problem 51E

To determine

To write: The first four terms of two different arithmetic sequence with common difference of $-3$ .Write an equation for the $n\mathrm{th}$ term of each arithmetic sequence.

Answer to Problem 51E

First four term of first sequence is $-6,-9,-12,-15$ , $a_n=-3-3n$ .

First four term of second sequence is $-1,-4,-7,-10$ , $a_n=2-3n$ .

Explanation of Solution

Given information: Common difference $=-3$ .

Calculation: Common difference $=-3$ .

Let first sequence is $-6,-9,-12,-\mathrm{15.....}$ .

Common difference $=a_2-a_1=-9+6=-3$ .

  $n\mathrm{th}$ Term of arithmetic sequence is $a_n=a+(n-1)d$

Here $a=-6\mathrm{and}d=-3$ put this value in $a_n$

  $\begin{array}{l}{a}_n=a+(n-1)d\\ a_n=-6+(n-1)(-3)\\ a_n=-6-3n+3\\ a_n=-3-3n\end{array}$

Let second sequence is $-1,-4,-7,-10,\mathrm{...}$

  $n\mathrm{th}$ Term of arithmetic sequence is $a_n=a+(n-1)d$

Here $a=-1\mathrm{and}d=-3$ put this value in $a_n$

  $\begin{array}{l}{a}_n=a+(n-1)d\\ a_n=-1+(n-1)(-3)\\ a_n=-1-3n+3\\ a_n=2-3n\end{array}$

Therefore First four term of first sequence is $-6,-9,-12,-15$ , $a_n=-3-3n$ .

First four term of second sequence is $-1,-4,-7,-10$ , $a_n=2-3n$ .