Chapter 12.3, Problem 39E

To determine

Find the third term of the geometric sequence having fourth term $a_4=\frac{5}{2}$ and common ratio $r=\frac{2}{5}$ .

Answer to Problem 39E

The fifth term of the geometric sequence is $a_3=\frac{25}{4}$ .

Explanation of Solution

Given information:

It is given in the question that, fourth term $a_4=\frac{5}{2}$ and common ratio $r=\frac{2}{5}$ .

Formula used:

  $a_n=a{r}^{n-1}$

Where,

  $\begin{array}{l}aisthe{1}^{st}term\\ nisthenumberofterm\\ risthecommonratio\end{array}$

Calculation:

It is given in the question that, fourth term $a_4=\frac{5}{2}$ and common ratio $r=\frac{2}{5}$ .

  $\begin{array}{l}r=\frac{2}{5}\\ a_4=\frac{5}{2}\Rightarrow a{r}^3=\frac{5}{2}\Rightarrow a=\frac{\frac{5}{2}}{r^3}\Rightarrow a=\frac{\frac{5}{2}}{{\left(\frac{2}{5}\right)}^3}\Rightarrow a=\frac{\frac{5}{2}}{\left(\frac{8}{125}\right)}\Rightarrow a=\frac{5}{2}*\frac{125}{8}\Rightarrow a=\frac{625}{16}\end{array}$

So, $a=\frac{625}{16}$ & $r=\frac{2}{5}$

By putting $a=\frac{625}{16}$ & $r=\frac{2}{5}$ & $n=5$ it in formula

  $\begin{array}{l}{a}_3=\frac{625}{16}*{\left(\frac{2}{5}\right)}^{3-1}\\ =\frac{625}{16}*{\left(\frac{2}{5}\right)}^2\\ =\frac{625}{16}*\frac{4}{25}\\ =\frac{25}{4}\end{array}$