Question

What is the 6th term of the geometric sequence where a1 = 128 and a3 = 8?

Answer 1

Answer:

The 6th term of the geometric sequence is, $0.125=\frac{1}{8}$

Step-by-step explanation:

The nth term of the geometric sequence is given by:

$a_n = a_1 \cdot r^{n-1}$

where,

$a_1$ is the first term

r is the common ratio of the term

n is the number of terms

As per the statement:

$a_1 = 128$ and $a_3 = 8$

For n = 3

$a_3 = a_1 \cdot r^2$

Substitute the given values we have;

$8 = 128 \cdot r^2$

Divide both sides by 128 we have;

$\frac{1}{16} = r^2$

⇒$\sqrt{\frac{1}{16}} =r$

⇒$\frac{1}{4} = r$

or

$r =\frac{1}{4}=0.25$

We have to find the 6th term of the geometric sequence.

For n = 6 we have;

$a_6 = a_1 \cdot r^5$

⇒$a_6 = 128 \cdot (0.25)^5 = 128 \cdot 0.0009765625$

Simplify:

$a_6 =0.125=\frac{1}{8}$

Therefore, the 6th term of the geometric sequence is, $0.125=\frac{1}{8}$

Answer 2

I have not learned this but if I were to guess...

a1 = 128
a2 = 32
a3 = 8
a4 = 2
a5 = 1/2
a6 = 1/8

Therefore I believe the 6th term is 1/8. The pattern is that it is dividing by 4.