Chapter 7.9, Problem 35PPS

To determine

To find the first and fourth terms of the given geometric sequence.

Answer to Problem 35PPS

First term is 9 and fourth term is $\frac{1}{3}$ .

Explanation of Solution

Given:

  $a_2=3\&a_3=1$

Concept used:

The n^th term of a geometric sequence with first term ‘a’ and common ratio ‘r’ is

  $a_n=a{r}^{n-1}$

Calculation:

  $\begin{array}{l}{a}_2=3\\ \Rightarrow ar=3\to \left(1\right)\end{array}$

&

  $\begin{array}{l}{a}_3=1\\ \Rightarrow a{r}^2=1\to \left(2\right)\end{array}$

Now dividing equation (2) by equation (1), we get

  $\begin{array}{l}\frac{a{r}^2}{ar}=\frac{1}{3}\\ \Rightarrow r=\frac{1}{3}\end{array}$

Now put $r=\frac{1}{3}$ in equation (1), we get

  $\begin{array}{l}a\left(\frac{1}{3}\right)=3\\ \Rightarrow a=9\end{array}$

Therefore first and fourth terms are

  $\begin{array}{l}\text{First}\text{term}=a=9\\ \text{Fourth}\text{term}=a_4=a{r}^3\\ =9{\left(\frac{1}{3}\right)}^3\\ =\frac{9}{27}\\ =\frac{1}{3}\end{array}$