Question

What is the 7th term of the geometric sequence where a1 = -625 and a2 = 125

Answer 1

Since the sequence is geometric, there is some constant $r$ such that the sequence is recursively given by

$a_n=ra_{n-1}$

By this definition, you can recursively substitute into the right hand side the definition for $a_{n-1},a_{n-2},\ldots$ to find an explicit formula for the $n$th term.

$a_n=ra_{n-1}=r^2a_{n-2}=r^3a_{n-3}=\cdots=r^{n-1}a_1=-625r^{n-1}$

You know the second term, which means you can find $r$:

$a_2=-625r^{2-1}\implies125=-625r\implies r=-\dfrac15$

So, the 7th term of the sequence is

$a_7=-625\left(-\dfrac15\right)^{7-1}=-\dfrac1{25}$