# 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 `=2112sqrt8`

# 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**

`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**

`-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**

`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 `=2112sqrt8`