# Given the two terms in a geometric sequence how do you find the recursive formula... a1=-4 and a4=-500?

Apr 27, 2016

Recursive formula is ${a}_{1} = - 4$ and ${a}_{n + 1} = 5 {a}_{n}$

#### Explanation:

In a geometric sequence, if $a$ is the first term and $r$ is the ratio between a term and its preceding term, then ${n}^{t h}$ term is given by $a \times {r}^{n - 1}$.

Here $a = {a}_{1} = - 4$ and as fourth term is $- 500$

$- 4 \times {r}^{4 - 1} = - 500$ or $4 {r}^{3} = 500$ or ${r}^{3} = 125$ and hence

$r = \sqrt{125} = \sqrt{5 \times 5 \times 5} = 5$

Hence recursive formula is ${a}_{1} = - 4$ and ${a}_{n + 1} = 5 {a}_{n}$

May 1, 2016

Recursive formulation:

$\left\{\begin{matrix}{a}_{1} = - 4 \\ {a}_{n + 1} = 5 {a}_{n} \textcolor{w h i t e}{000} \left(n = 1 2 3 \ldots\right)\end{matrix}\right.$

#### Explanation:

We are given:

$\left\{\begin{matrix}{a}_{1} = - 4 \\ {a}_{4} = - 500\end{matrix}\right.$

The general formula for the $n$th term of a geometric series is:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ is the common ratio.

A recursive formula can be given as:

$\left\{\begin{matrix}{a}_{1} = a \\ {a}_{n + 1} = r {a}_{n} \textcolor{w h i t e}{000} \left(n = 1 2 3 \ldots\right)\end{matrix}\right.$

In our example:

${5}^{3} = 125 = \frac{- 500}{- 4} = {r}_{4} / {r}_{1} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{a}}} {r}^{4 - 1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{a}}} {r}^{1 - 1}} = {r}^{3}$

So the only possible Real value for $r$ is $\sqrt{{5}^{3}} = 5$.

$\textcolor{w h i t e}{}$
Footnote

There are two other possibilities for a geometric sequence with ${a}_{1} = - 4$ and ${a}_{4} = - 500$, which are sequences of Complex numbers.

This is because ${5}^{3}$ has two other cube roots, namely $5 \omega$ and $5 {\omega}^{2}$, where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$. Either of these will also work as a suitable common ratio.