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

Question

1529 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T13:51:59

The general term is $t_n = 6 xx (-5)^(n - 1)$ .

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

$-30 = a xx r^(2 - 1)$

$3750 = a xx r^(5 - 1)$

Solve by substitution.

$-30 = a xx r$

$a = -30/r$

$3750 = -30/r xx r^4$

$3750 = -30r^3$

$-125 = r^3$

$r = -5$

$:.-30 = a xx (-5)^1$

$-30 = -5a$

$a = 6$

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

The general term is given by $t_n = a xx r^(n - 1)$

$t_n = 6 xx (-5)^(n - 1)$

Hopefully this helps!

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