# How do you write an nth term rule for a_2=-30 and a_5=3750?

Sep 7, 2016

The general term is ${t}_{n} = 6 \times {\left(- 5\right)}^{n - 1}$.

#### Explanation:

Write a system of equations in terms of $a$ and $d$.

$- 30 = a \times {r}^{2 - 1}$

$3750 = a \times {r}^{5 - 1}$

Solve by substitution.

$- 30 = a \times r$

$a = - \frac{30}{r}$

$3750 = - \frac{30}{r} \times {r}^{4}$

$3750 = - 30 {r}^{3}$

$- 125 = {r}^{3}$

$r = - 5$

$\therefore - 30 = a \times {\left(- 5\right)}^{1}$

$- 30 = - 5 a$

$a = 6$

$\therefore$The first term is $6$ and the common ratio is $- 5$.

The general term is given by ${t}_{n} = a \times {r}^{n - 1}$

${t}_{n} = 6 \times {\left(- 5\right)}^{n - 1}$

Hopefully this helps!