Chapter 7.5, Problem 39E

Solution

To state: The recursive rule for the sequence $-3,-2,5,-3,-2,\dots$ .

The resultant recursive rule is $a_n=\left(-1\right)\left(a_{n-2}+a_{n-1}\right)$ .

Given information:

The given sequence is $-3,-2,5,-3,-2,\dots$ .

Explanation:

Consider the given sequence,

  $-3,-2,5,-3,-2,\dots$

The first term of the sequence is: $a_1=-3$

The common difference ( d ) is:

  $\begin{array}{c}-2-\left(-3\right)=1\\ 5-\left(-3\right)=8\end{array}$

The given sequence is not an arithmetic sequence because the common difference of two successive terms is not equal.

Now check if the sequence is geometric.

The common ratio ( r ) is:

  $\begin{array}{c}r=\frac{-2}{-3}\\ =0.667\\ r=\frac{5}{-2}\\ =-2.5\end{array}$

The given sequence is not a geometric sequence because the common ratio is not same from the two terms.

Now again consider the sequence $-3,-2,5,-3,-2,\dots$ ,

Beginning with the third term in the sequence, $a_3$ can be obtained by multiplying the sum of the first and second terms by -1.

  $\begin{array}{c}{a}_3=\left(-1\right)\left(a_1+a_2\right)\\ =\left(-1\right)\left(-3+\left(-2\right)\right)\\ =\left(-1\right)\left(-5\right)\\ =5\end{array}$

Similarly, the fourth term can be obtained by multiplying the sum of the second and third terms by -1.

  $\begin{array}{c}{a}_4=\left(-1\right)\left(a_2+a_3\right)\\ =\left(-1\right)\left(-2+5\right)\\ =\left(-1\right)\left(3\right)\\ =-3\end{array}$

So the recursive rule for the sequence $a_1=-3,a_2=-2$ is $a_n=\left(-1\right)\left(a_{n-2}+a_{n-1}\right)$ .