Question

Find the 10th term of the geometric sequence whose common ratio is 1/2 and whose 1st term is 2.

Answer 1

For an nth term in a geometric sequence

`U(n) = a ({r})^(n - 1)`

where n is the number of terms

r is the common ratio

a is the first term

From the question

a = 2

r = 1/2

n = 10

So the 10th term of the sequence is

`U(10) = 2 ({ (1)/(2) })^(10 - 1) \\ \\ = 2 ({ (1)/(2) })^(9) \\ \\ \\ = (1)/(256)`

Hope this helps you

Answer 2

`(1)/(256)`

Explanation:

Geometric sequence means there is a common ratio. All that means is term divided previous term is the same across your sequence.

ONE WAY:

So we are given here that:

`(f(2))/(f(1))=(1)/(2)` and that the first term which is
`f(1)` is 2.

`(f(2))/(2)=(1)/(2)`

This implies
`f(2)=1` after multiplying both sides by 2 and getting that
`f(2)=(1)/(2)(2)=(2)/(2)=1`.

So you have that

2,1,...

basically you can just multiply by 1/2 to keep generating more terms of the sequence.

Third term would be
`f(3)=1((1)/(2))=(1)/(2)`.

Fourth term would be
`f(4)=(1)/(2)((1)/(2))=(1)/(4)`.

...keep doing this til you get to the 10th term.

ANOTHER WAY:

Let's make a formula.

`f(n)=ar^(n-1)`

`a` is the first term.

`r` is the common ratio.

And we want to figure out what happens at
`n=10`.

Let's plug in our information we have

`a=2`

`r=(1)/(2)`:

`f(10)=2((1)/(2))^(10-1)`

Put into calculator or do by hand...

`f(10)=2((1)/(2))^9`

`f(10)=2((1^9)/(2^9))`

`f(10)=2((1)/(2^9))`

`f(10)=(2)/(2^9)`

`f(10)=(2)/(2(2^8))`

`f(10)=(1)/(2^8)`

Scratch work:

`2^8=2^5 \cdot 2^3=32 \cdot 8=256`.

End scratch work.

The answer is that the tenth term is
`(1)/(256)`