# How do you write an nth term rule for a_2=-20 and a_4=-5?

Jan 11, 2017

${a}_{n} = - \frac{40}{2} ^ \left(n - 1\right)$ or ${a}_{n} = 40 {\left(- \frac{1}{2}\right)}^{n - 1}$

#### Explanation:

Although it is not specifically mentioned as the question is framed under "Geometric Series", it is assumed to be so.

In a geometric series if firs term is ${a}_{1}$ and common ratio is $r$

the ${n}^{t h}$ term ${a}_{n} = {a}_{1} \times {r}^{n - 1}$

As ${a}_{2} = - 20$ and ${a}_{4} = - 5$, we have

${a}_{1} \times r = - 20$ .................(1) $-$ and ${a}_{1} = - \frac{20}{r}$

${a}_{1} \times {r}^{3} = - 5$ .................(2)

Dividing (2) by (1), we get ${r}^{2} = \frac{- 5}{- 20} = \frac{1}{4}$

Hence $r = \pm \frac{1}{2}$

If $r = \frac{1}{2}$, ${a}_{1} = - \frac{20}{\frac{1}{2}} = - 40$ and ${a}_{n} = - 40 \times {\left(\frac{1}{2}\right)}^{n - 1} = - \frac{40}{2} ^ \left(n - 1\right)$

and if $r = - \frac{1}{2}$, ${a}_{1} = - \frac{20}{- \frac{1}{2}} = 40$ and ${a}_{n} = 40 {\left(- \frac{1}{2}\right)}^{n - 1} = 40 {\left(- \frac{1}{2}\right)}^{n - 1}$