Chapter 12.5, Problem 53E

To determine

To find: the first four terms of geometric sequence.

Answer to Problem 53E

  $\begin{array}{l}{a}_1=8\sqrt{2}\\ a_2=-16\\ a_3=16\sqrt{2}\\ a_4=-32\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{c}{a}_5=32\sqrt{2}\\ r=-\sqrt{2}\end{array}$

Calculation:

The general form of geometric series is $a_n=a_1{r}^{n-1}$ . Now substitute $n=5$ into the equation to evaluate first term of geometrical series as shown below.

  $\begin{array}{c}{a}_5=a_1{r}^{5-1}\\ 32\sqrt{2}=a_1{r}^4\left[\text{Put }{a}_5\right]\\ a_1=\frac{32\sqrt{2}}{r^4}\\ =\frac{32\sqrt{2}}{{\left(-\sqrt{2}\right)}^4}\left[\text{Put }r\right]\\ =8\sqrt{2}\end{array}$

Now put $n=2,3,4$ and evaluate the other three terms of the geometric series as follows:

  $\begin{array}{c}{a}_2=a_1{r}^{2-1}\\ =\left(8\sqrt{2}\right){\left(-\sqrt{2}\right)}^1\left[\text{Put }{a}_1\text{ and }r\right]\\ =-16\\ a_3=a_2{r}^{3-2}\\ =-16{\left(-\sqrt{2}\right)}^1\left[\text{Put }{a}_2\text{ and }r\right]\\ =16\sqrt{2}\\ a_4=a_3{r}^{4-3}\\ =16\sqrt{2}{\left(-\sqrt{2}\right)}^1\left[\text{Put }{a}_3\text{ and }r\right]\\ =-32\end{array}$

Thus, the first four terms of the geometrical series are $8\sqrt{2},-16,16\sqrt{2}\text{ and}-32$ .