Chapter 8.3, Problem 36E

To determine

Write a rule for the nth term of the geometric sequence.

Answer to Problem 36E

Rule for the nth term of the geometric sequence is $\boxed{a_n=-2\cdot {\left(5\right)}^{n-1}}$

Explanation of Solution

Given information:

  $\begin{array}{l}{a}_2=-10\\ a_6=-6250\end{array}$

Formula used:

Nth term formula of a GP series $a_n=a_1{r}^{n-1}$

Write a system of equation using formula $a_n=a_1{r}^{n-1}$:

Equation 1:

  $\begin{array}{l}{a}_2=a_1{r}^{2-1}\\ -10=a_{1}r\end{array}$

Equation 2:

  $\begin{array}{l}{a}_6=a_1{r}^{6-1}\\ -6250=a_1{r}^5\end{array}$

Solve Equation 1 for $a_1$ :

  $a_1=\frac{-10}{r}$

Solve Equation 2 for $a_1$ :

  $a_1=\frac{-6250}{r^5}$

Now equating $\frac{-10}{r}$ and $\frac{-6250}{r^5}$ as they are equivalent to $a_1$ :

  $\begin{array}{l}\frac{-10}{r}=\frac{-6250}{r^5}\\ \\ \frac{r^5}{r}=\frac{-6250}{-10}\\ \\ r^4=625\\ \\ \boxed{r=5}\end{array}$

Now putting value of $r$ in Equation 1 to find the value of $a_1$ :

  $\begin{array}{l}-10=a_{1}r\\ -10=a_1\left(5\right)\\ a_1=\frac{-10}{5}\\ \boxed{a_1=-2}\end{array}$

Putting value of $a_1$ and $r$ in $a_n=a_1{r}^{n-1}$ to find rule for nth term:

  $\boxed{a_n=-2\cdot {\left(5\right)}^{n-1}}$