Question

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

Answer 1

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.

Answer 2

The 6th term of the geometric sequence is,
`0.125=(1)/(8)`

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;

`(1)/(16) = r^2`

⇒
`\sqrt{(1)/(16)} =r`

⇒
`(1)/(4) = r`

or

`r =(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=(1)/(8)`

Therefore, the 6th term of the geometric sequence is,
`0.125=(1)/(8)`