# Finding the nth term

## In a nutshell

Sequences defined using an $n^{th}$ term formula help us find the sequence from the position of the terms in the sequence. The $n^{th}$ term method is more useful, as if you want to find the $100^{th}$ term, it is not necessary to find any of the previous terms. You should be able to find the $n^{th}$ term formula for a linear or a quadratic sequence.

## Generating a sequence

The $n^{th}$ term formula can be used to generate a sequence. Start by substituting $n=1$ into the formula to find the first term. Then substitute $n=2$ for the second term, $n=3$ for the third term and so on.

##### Examples

**NTH TERM FORMULA** | **SEQUENCE** |

$4n+4$ | $8,12, 16, 20,24$ |

$8-6n$ | $2, -4, -10, -16, -22$ |

$n^2+2$ | $3, 6, 11, 18, 27$ |

## Find the nth term formula for a linear sequence

A linear or arithmetic sequence is one where the difference between adjacent terms are always the same. The term-to-term rule would be to $+$ or $-$ the same number each time.

It is possible to find the $n^{th}$ term formula for an arithmetic sequence.

##### Example 1

*Find the* $n^{th}$ *term formula for the sequence* $2, 5, 8, 11$.

*Start by putting the sequence, and their positions* $n$ *into a table. Work out the difference each time, in this case the difference is* $3$, *so add another row in the table for* $3n$. *Then think about how to get from *$3n$ *to the sequence, in this case each term in our sequence is* $1$ *less than the* $3n$ *row.*

$\begin {array} {c|c c c c c c c}n(position) & 1 && 2 && 3 &&4 \\ \hline n^{th} \space term & 2 && 5 && 8 && 11 \\difference && \xrightarrow[+3]{} && \xrightarrow[+3]{} &&\xrightarrow[+3]{} \\3n & 3 && 6 && 9 && 12 \\3n-1 & 2 && 5 && 8 && 11\end {array}$

$\underline {n^{th} \space term = 3n-1}$

**Note***:** Always check the formula by substituting values for *$n$* to see if it works.*

## Finding the nth term for a quadratic sequence - higher only

A quadratic sequence has its $n^{th}$ term formula in quadratic form as

$n^{th} \space term = an^2+bn+c$

where $a, b$ and $c$ are constants to be found. You can find the $n^{th}$ term formula by putting the sequence and their positions $n$ into a table. A quadratic sequence can be identified by looking at adjacent terms in the sequence. If the difference between terms goes up in equal steps, then it is a quadratic sequence.

##### Example 2

$\begin{array}{c c c c c c c c c}5 & & 8 & & 13 & & 20 & & 29 \\& \xrightarrow[+3]{} && \xrightarrow[+5]{} && \xrightarrow[+7]{} && \xrightarrow[+9]{} \end{array}$

*The adjacent terms go up in increasing steps, so this is a quadratic sequence.*

#### PROCEDURE

$1.$ | Fill the sequence in a table with their positions, $n$, in the first row, and the sequence, $n^{th}$ term, in the second row. |

$2.$ | Calculate the first row of differences by calculating the difference between adjacent terms. Name this row $d1$. |

$3.$ | Calculate the second row of differences by calculating the difference between the numbers in $d1$ row. Name this row $d2$. |

$4.$ | Make the first number in the row $d2$ equal to $2a$. Solve this equation for $a$. |

$5.$ | Make the first number in the row $d1$ equal to $3a+b$. Use the solution for $a$ from step $4$ in the equation and solve the equation for $b$. |

$6.$ | Make the first term in the sequence equal to $a+b+c$. Use the solutions for $a$ and $b$ in the equation and solve to find $c$. |

$7.$ | Substitute the values of $a, b$ and $c$ into $n^{th} \space term = an^2+bn+c$ to obtain the formula for the $n^{th}$ term. |

##### Example 3

*Find the *$n^{th}$* term formula for the sequence*

$5, 11, 19, 29, 41$

*Start by filling the rows in a table with the first and second row of differences*

$\begin{array}{c | c c c c c c c c c}n &1 && 2 && 3 && 4 &&5 \\ \hline n^{th} \space term & 5 && 11 && 19 && 29 && 41 \\d1 && +6 && +8 &&+10 && +12 \\d2 &&&+2 &&+2 && +2\end{array}$

*Form the equations from the table.*

$\begin{array}{c | c c c c c c c c c}n &1 && 2 && 3 && 4 &&5 \\ \hline n^{th} \space term & \underbrace{5}_\text{a+b+c} && 11 && 19 && 29 && 41 \\d1 && \underbrace{+6}_\text{3a+b} && +8 &&+10 && +12 \\d2 &&&\underbrace{+2}_\text{2a} &&+2 && +2 \\\end{array}$

*$\begin {aligned}2a&=2 \\3a+b&=6 \\a+b+c&=5\end {aligned}$*

*Solve the equations for *$a, b$* and *$c$*.*

$\begin {aligned}2a&=2 \\a&=1 \\ \\3a+b&=6 \\3 \times 1 + b &=6 \\3+b &= 6\\b&= 3 \\ \\a+b+c&=5 \\1+ 3 + c &=5 \\4 +c &=5 \\c &=1\end {aligned}$

*Substitute the values of *$a, b$* and *$c$* into the formula *$n^{th} \space term = an^2+bn+c$* to find the formula.*

$\underline {n^{th}\space term = n^2+3n+1}$