Question

How do you write the ${\text{n}}^{\text{th}}$ term rule for the sequence $1,3,5,7,9,....$?

Answer 1

To find the ${\text{n}}^{\text{th}}$ term rule for the given sequence:

Consider the given sequence.

$1,3,5,7,9,....$

Since this is a arithmetic sequence, so the first term is $1$ and the common difference is calculated as:

$3-1=2$

Using the ${\text{n}}^{\text{th}}$ arithmetic sequence formula, ${\text{a}}_{\text{n}}=\text{a}+\left(\text{n}-1\right)\text{d}$, where ${\text{a}}_{\text{n}}$ is ${\text{n}}^{\text{th}}$ terms, $\text{a}$ is first term, $\text{n}$ is number of terms and $\text{d}$ is a common difference.

$\begin{array}{rcl}{\text{a}}_{\text{n}}& =& \text{a}+\left(\text{n}-1\right)\text{d}\left[{\text{n}}^{\text{th}}\text{term formula}\right]\\ & \Rightarrow & {\text{a}}_{\text{n}}=1+\left(\text{n}-1\right)2\left[\text{Substitute a=1, d=2}\right]\\ & \Rightarrow & {\text{a}}_{\text{n}}=1+2\text{n}-2\left[\text{Simplify}\right]\\ \therefore {\text{a}}_{\text{n}}& =& 2\text{n}-1\end{array}$

Hence, the ${\text{n}}^{\text{th}}$ term rule is $\begin{array}{rcl}{\mathbf{\text{a}}}_{\mathbf{\text{n}}}& \mathbf{=}& \mathbf{2}\mathbf{\text{n}}\mathbf{-}\mathbf{1}\end{array}$.