Formula for `n^(th)` term of a sequence - multiplication
In order to predict the `n^(th)` term of a sequence you will need to create a formula.
For sequence patterns of geometric progressions or geometric sequences (or multiplications) this is worked out by using the formula.
`ar^(n-1)`
Where
`a=` first term
`r=` the multiple
`n=n^(th)` number
We can help you remember this as follows
Arrrr `(ar)` not one `(n-1)` of us
`ar^(n-1)`
Example 1
We know the following sequence is a geometric sequence but what is the formula of the `n^(th)` term and what is the `8^(th)` term.
`2` | `6` | `18` | `54` | `162` |
We can check that it is a multiple sequence by dividing each term by the previous term. | ||||
`6/2=3` | `18/6=3` | `54/18=3` | `162/54=3` |
We have a multiple of `3`
So the formula for this is
`ar^(n-1)`
So here
`a=` first term `=2`
`r=` the multiple `=3`
So the formula is
`2times3^(n-1)`
Now we need to check the formula is correct.
Try different values of `n` in the formula `2times3^(n-1)`
`n=1` term `=2times3^(1-1)=2times3^0=2times1=2`
`n=2` term `=2times3^(2-1)=2times3^1=2times3=6`
`n=3` term `=2times3^(3-1)=2times3^2=2times9=18`
`n=4` term `=2times3^(4-1)=2times3^3=2times27=54`
This is correct
The 8th term would be
If `n=8` then `2times3^(n-1)=2times3^(8-1)=2times3^7=4374`
The 8th term `=4,374`
Example 2
We know the following sequence is a geometric sequence but what is the formula of the `n^(th)` term and what is the `8^(th)` term.
`-2` | `4` | `-8` | `16` | `-32` | `64` |
We can check that it is a multiple sequence by dividing each term by the previous term. | |||||
`4/-2=-2` | `(-8)/4=-2` | `16/-8=-2` | `(-32)/16=-2` | `64/-32=-2` |
We have a multiple of `-2`
So the formula for this is
`ar^(n-1)`
So here
`a=` first term `=-2`
`r=` the multiple `=-2`
So the formula is
`-2times(-2)^(n-1)`
Now we need to check the formula is correct.
Try different values of `n` in the formula `-2times(-2)^(n-1)`
`n=1` term `=-2times(-2)^(1-1)=-2times(-2)^0=-2times1=-2`
`n=2` term `=-2times(-2)^(2-1)=-2times(-2)^1=-2times-2=4`
`n=3` term `=-2times(-2)^(3-1)=-2times(-2)^2=-2times4=-8`
`n=4` term `=-2times(-2)^(4-1)=-2times(-2)^3=-2times(-8)=16`
This is correct
The 8th term would be
If `n=8` then `-2times(-2)^(n-1)=-2times(-2)^(8-1)=-2times(-2)^7`
`=-2times(-2xx-2xx-2xx-2xx-2xx-2xx-2)`
`=-2times(-128)=256`
Answer `=256`
Example 3
We know the following sequence is a geometric sequence but what is the formula for the `n^(th)` term and what is the `8^(th)` term.
`40` | `20` | `10` | `5` | `2.5` |
We can check that it is a multiple sequence by dividing each term by the previous term. | ||||
`20/40=0.5` | `10/20=0.5` | `5/10=0.5` | `2.5/5=0.5` |
We have a multiple of `0.5`
So the formula for this is
`ar^(n-1)`
So here
`a=` first term `=40`
`r=` the multiple `=0.5`
So the formula is
`40times(0.5)^(n-1)`
Now we need to check the formula is correct.
Try different values of `n` in the formula `40times(0.5)^(n-1)`
`n=1` term `=40times(0.5)^(1-1)=40times0.5^0=40times1=40`
`n=2` term `=40times(0.5)^(2-1)=40times0.5^1=40times0.5=20`
`n=3` term `=40times(0.5)^(3-1)=40times0.5^2=40times0.5xx0.5=10`
`n=4` term `=40times(0.5)^(4-1)=40times0.5^3=40times0.5xx0.5xx0.5=5`
This is correct
The 8th term would be
If `n=8` then `40times(0.5)^(n-1)=40times(0.5)^(8-1)=40times0.5^7`
`=40times(0.5xx0.5xx0.5xx0.5xx0.5xx0.5xx0.5)`
`=0.3125`
Answer the 8th term `=0.3125`
Example 4
We know the following sequence is a geometric sequence but what is the formula of the `n^(th)` term and what is the `8^(th)` term.
`3` | `3sqrt8` | `24` | `24sqrt8` |
We change to
`3` | `3sqrt8` | `3sqrt8sqrt8` | `3sqrt8sqrt8sqrt8` |
We can check that it is a multiple sequence by dividing each term by the previous term. | |||
`(3sqrt8)/3=sqrt8` | `(3sqrt8sqrt8)/(3sqrt8)=sqrt8` | `(3sqrt8sqrt8sqrt8)/(3sqrt8sqrt8)=sqrt8` |
We have a multiple of `sqrt8`
So the formula for this is
`ar^(n-1)`
So here
`a=` first term `=3`
`r=` the multiple `=sqrt8`
So the formula is
`3sqrt8^(n-1)`
Now we need to check the formula is correct.
Try different values of `n` in the formula `3sqrt8^(n-1)`
`n=1` term `=3sqrt8^(1-1)=3sqrt8^0=3times1=3`
`n=2` term `=3sqrt8^(2-1)=3sqrt8^1=3sqrt8`
`n=3` term `=3sqrt8^(3-1)=3sqrt8^2=3sqrt8sqrt8`
`n=4` term `=3sqrt8^(4-1)=3sqrt8^3=3sqrt8sqrt8sqrt8`
This is correct
The 8th term would be
Answer the 8th term `=1536sqrt8`