Question

What is the 6th term of the Geometric Sequence where a1 = 128 and a3=8?

Explicit formula:
a_n(r)^n-1

Please can you explain step-by-step to me how to find this?

Answer 1

The 6th term of the sequence is 0.125.

Explanation:

Given : Geometric Sequence where and

To find : The 6th term.

Solution :

The formula of nth term of geometric sequence is

........[1]

We have given,

To find ratio we put n=3 in [1]

Now, we know first term and common ratio

Put n=6 in [1] to get 6th term of the sequence is

Therefore, The 6th term of the sequence is 0.125.

Answer 2

The 6th term of the sequence is 0.125.

Explanation:

Given : Geometric Sequence where
`a_1=128` and
`a_3=8`

To find : The 6th term.

Solution :

The formula of nth term of geometric sequence is

`a_n=a_1* r^(n-1)` ........[1]

We have given,

`a_1=128`

`a_3=8`

To find ratio we put n=3 in [1]

`a_3=a_1* r^(3-1)`

`8=128* r^(2)`

`(8)/(128)= r^(2)`

`0.0625= r^(2)`

`r=√(0.0625)`

`r=0.25`

Now, we know first term
`a_1=128` and common ratio
`r=0.25`

Put n=6 in [1] to get 6th term of the sequence is

`a_6=a_1* r^(6-1)`

`a_6=128* (0.25)^(5)`

`a_6=128* 0.000976`

`a_6=0.125`

Therefore, The 6th term of the sequence is 0.125.