Chapter 6.6, Problem 55E

To determine

To write: the first 5 terms of the sequence $a_1=5$ , $a_n=3a_{n-1}+4$ and determine whether the sequence is an arithmetic sequence or the geometric sequence or neither.

Answer to Problem 55E

  $\begin{array}{l}{a}_1=5\\ a_2=19\\ a_3=61\end{array}$

  $\begin{array}{l}{a}_4=187\\ a_5=565\end{array}$

The given sequence is neither an arithmetic sequence nor a geometric sequence.

Explanation of Solution

Concept Used:

In Arithmetic sequence difference between each pair of consecutive terms is same.

In Geometric sequence the ratio between each pair of consecutive terms is same.

The first 5 terms of the given sequence are found by substituting $n=2,3,4,5$ in the formula $a_n=3a_{n-1}+4$ :

  $a_1=5$

  $\begin{array}{c}{a}_2=3a_{2-1}+4\\ =3\left(5\right)+4\\ =15+4\\ =19\end{array}$

  $\begin{array}{c}{a}_3=3a_2+4\\ =3\left(19\right)+4\\ =57+4\\ =61\end{array}$

  $\begin{array}{c}{a}_4=3a_3+4\\ =3\left(61\right)+4\\ =183+4\\ =187\end{array}$

  $\begin{array}{c}{a}_5=3a_4+4\\ =3\left(187\right)+4\\ =561+4\\ =565\end{array}$

First find the difference between each pair of consecutive terms:

  $\begin{array}{c}19-5=14\\ 61-19=42\\ 187-61=126\\ 565-187=378\end{array}$

Now, find the ratio between each pair of consecutive terms:

  $\begin{array}{c}\frac{19}{5}=\frac{19}{5}\\ \frac{61}{19}=\frac{61}{19}\\ \frac{187}{61}=\frac{187}{61}\\ \frac{565}{187}=\frac{565}{187}\end{array}$

There is neither a common difference nor a common ratio between the terms.

So, the sequence is neither an arithmetic sequence nor a geometric sequence.