# Triangles and the `n^(th)` term

`1` `3` `6` `10` `15`

You can recognise a triangular sequence by picturing it as follows:

We can see a pattern emerge

`1^(st)` | term `=1` | `=1` |

`2^(nd)` | term `=1+2` | `=3` |

`3^(rd)` | term `=1+2+3` | `=6` |

`4^(th)` | term `=1+2+3+4` | `=10` |

`5^(th)` | term `=1+2+3+4+5` | `=15` |

This also gives us a clue as to how we find the `n^(th)` term of any triangular sequence (But note the following only applies when the sequence starts at the number 1).

## At first, just take one large even set of `n^(th)` terms

Try `12^(th)` term

The `12^(th)` term would be

`1+2+3+4+5+6+7+8+9+10+11+12=78`

But a mathematician found that if you started by adding the `1^(st)` and last term together and then continue this arrangement then a pattern emerges.

i.e.

`12^(th)` term`=` | `1+12` | `=13` | |

`2+11` | `=13` | ||

`3+10` | `=13` | ||

`4+9` | `=13` | ||

`5+3` | `=13` | ||

`6+7` | `=13` | ||

`6times13` |
`=78` |

So the pattern is `13` and the multiple is `6`.

This works for any even numbered triangular number.

Try `16^(th)` term

`16^(th)` term`=` | `1+16` | `=17` | |

`2+15` | `=17` | ||

`3+14` | `=17` | ||

`4+13` | `=17` | ||

`5+12` | `=17` | ||

`6+11` | `=17` | ||

`7+10` | `=17` | ||

`8+9` | `=17` | ||

`8times17` |
`=136` |

So the pattern is `17` and the multiple is `8`.

Now we can form a formula for this

### Stage 1 find the pattern number

`12^(th)` term `=(12+1)=13`

`16^(th)` term `=(16+1)=17`

So the pattern number `=(n^(th)\ \n\u\m\b\e\r+1)`

### Stage 2

And the amount of times we repeat this pattern number (the multiple)

`12^(th)` term `=(12)/2=6`

`16^(th)` term `=(16)/2=8`

So the multiple `=(n^(th)\ term)/2`

Put stage 1 and stage 2 together we get:

`(n^(th)\ term+1)times(n^(th)\ term)/2`

So the `n^(th)\ term=(n+1)timesn/2`

And amazingly this also works for odd triangular numbers too.

**Example 1**

What is the `9^(th)` term of the triangular sequence

Either draw it

`9^(th)` term`=` | `1+9` | `=10` | |

`2+8` | `=10` | ||

`3+7` | `=10` | ||

`4+6` | `=10` | ||

`4times10` | `+5=45` |

**NOTE:**

We add `5` because that is the number missed in the series of semi-circles above.

or using the formula

`(n+1)timesn/2`

`9^(th)` term | `=(9+1)times9/2` |

`=(10times9)/2` | |

`=5times9` | |

`=45` |

**Example 2**

What is the `11^(th)` term of the triangular sequence

Either draw it

`11^(th)` term`=` | `1+11` | `=12` | |

`2+10` | `=12` | ||

`3+9` | `=12` | ||

`4+8` | `=12` | ||

`5+7` | `=12` | ||

`5times12` | `+6=66` |

**NOTE:**

We add `6` because that is the number missed in the series of semi-circles above.

or using the formula

`(n+1)timesn/2`

`11^(th)` term | `=(11+1)times11/2` |

`=(12times11)/2` | |

`=6times11` | |

`=66` |

**Example 3**

What is the `27^(th)` term of the triangular sequence

We are not going to draw this out but we can just rely on the formula.

`(n+1)timesn/2`

`27^(th)` term | `=(27+1)times27/2` |

`=(28times27)/2` | |

`=14times27` | |

`=378` |

**Example 4**

What is the `28^(th)` term of the triangular sequence

Again we are not going to draw this out but the formula gives:

`(n+1)timesn/2`

`28^(th)` term | `=(28+1)times28/2` |

`=(29times28)/2` | |

`=29times14` | |

`=406` |

(As a point of interest comparing example 3 and the answer 378 and here 406 the difference is 28 which would be correct.)

# Triangles and the `n^(th)` term

`1` `3` `6` `10` `15`

You can recognise a triangular sequence by picturing it as follows:

We can see a pattern emerge

`1^(st)` | term `=1` | `=1` |

`2^(nd)` | term `=1+2` | `=3` |

`3^(rd)` | term `=1+2+3` | `=6` |

`4^(th)` | term `=1+2+3+4` | `=10` |

`5^(th)` | term `=1+2+3+4+5` | `=15` |

This also gives us a clue as to how we find the `n^(th)` term of any triangular sequence (But note the following only applies when the sequence starts at the number 1).

## At first, just take one large even set of `n^(th)` terms

Try `12^(th)` term

The `12^(th)` term would be

`1+2+3+4+5+6+7+8+9+10+11+12`

But a mathematician found that if you started by adding the `1^(st)` and last term together and then continue this arrangement then a pattern emerges.

i.e.

`12^(th)` term`=` | `1+12` | `=13` |

`2+11` | `=13` | |

`3+10` | `=13` | |

`4+9` | `=13` | |

`5+3` | `=13` | |

`6+7` | `=13` | |

`6times13` |

This works for any even numbered triangular number.

Try `16^(th)` term

`16^(th)` term`=` | `1+16` | `=17` |

`2+15` | `=17` | |

`3+14` | `=17` | |

`4+13` | `=17` | |

`5+12` | `=17` | |

`6+11` | `=17` | |

`7+10` | `=17` | |

`8+9` | `=17` | |

`8times17` |

Now we can form a formula for this

### Stage 1

`12th` triangular number `=(12+1)=13`

`16th` triangular number `=(16+1)=17`

So the pattern number `=(nth\ \n\u\m\b\e\r+1)`

We can see that a TOTAL of one set `=(n^(th)\ \ term+1)`

as above this was `12^(th)` term `=(12+1)=13`

and `16^(th)` term `=(16+1)=17`

### Stage 2

And the amount of times we repeat this set would be:

How many sets `=n^(th)div2` or `(n^(th)\ term)/2`

as above this was `12^(th)` term `=(12)/2=6`

and `16^(th)` term `=(16)/2=8`

Put stage 1 and stage 2 together we get:

`(n^(th)\ term+1)times(n^(th)\ term)/2`

or `(n+1)timesn/2`

And amazingly this also works for odd numbers of terms too.

**Example 1**

What is the `9^(th)` term of the triangular sequence

Either draw it

`9^(th)` term`=` | `1+9` | `=10` | |

`2+8` | `=10` | ||

`3+7` | `=10` | ||

`4+6` | `=10` | ||

`4times10` | `+5=45` |

**NOTE:**

We add `5` because that is the number missed in the series of semi-circles above.

or using the formula

`(n+1)timesn/2`

`9^(th)` term`=` | `=(9+1)times9/2` |

`=(10times9)/2` | |

`=5times9` | |

`=45` |

**Example 2**

What is the `11^(th)` term of the triangular sequence

Either draw it

`11^(th)` term`=` | `1+11` | `=12` | |

`2+10` | `=12` | ||

`3+9` | `=12` | ||

`4+8` | `=12` | ||

`5+7` | `=12` | ||

`5times12` | `+6=66` |

**NOTE:**

We add `6` because that is the number missed in the series of semi-circles above.

or using the formula

`(n+1)timesn/2`

`11^(th)` term`=` | `=(11+1)times11/2` |

`=(12times11)/2` | |

`=6times11` | |

`=66` |

**Example 3**

What is the `27^(th)` term of the triangular sequence

We are not going to draw this out but we can just rely on the formula.

`(n+1)timesn/2`

`27^(th)` term`=` | `=(27+1)times27/2` |

`=(28times27)/2` | |

`=14times27` | |

`=378` |

**Example 4**

What is the `28^(th)` term of the triangular sequence

Again we are not going to draw this out but the formula gives:

`(n+1)timesn/2`

`28^(th)` term`=` | `=(28+1)times28/2` |

`=(29times28)/2` | |

`=29times14` | |

`=406` |

(As a point of interest comparing example 3 and the answer 378 and here 406 the difference is 28 which would be correct.)