Question

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

Answer 1

Answer:

The nth term of the geometric sequence is given by;

$a_n = a_1r^{n-1}$, .....[1] n is the number of terms.

where,

$a_1$ is the first term and r is the common ratio of the terms.

As per the statement:

Given:

$a_1= 625$ and $a_2 = -125$

Using [1];

$a_2 = a_1 \cdot r$

Substitute the value of $a_1$ we have;

$-125 = 625 \cdot r$

Divide both sides by 625 we have;

$-\frac{1}{5} = r$

or

$r= -\frac{1}{5}$

We have to find 7th term of the geometric sequence.

For n = 7 we have;

$a_7 = a_1 \cdot r^6$

Substitute the given values we have;

$a_7 = 625 \cdot (-\frac{1}{5})^6 = \frac{625}{15625} = \frac{1}{25} =0.04$

therefore, the 7th term of the geometric sequence is, 0.04

Answer 2

Common ratio, r = (-125)/625 = -0.2
First term, a = 625

7 term
= ar^6
= 625(-0.2)^6
= -0.04