# How do you write a rule for the nth term 7,5,3,1,-1?

Jun 27, 2018

each term is going down by 2

$\implies - 2 n$

this gives -2, -4, -6, -8

to get the sequence we need to find the adjuster

opposite of -2 is +2, 7+2=9

$- 2 n + 9$ or $9 - 2 n$

Jun 27, 2018

${a}_{n} = 7 - 2 n$

#### Explanation:

First of all, notice that this is an arithmetic sequence, i.e. two consecutive terms differ by a common difference. In this case, two consecutive terms always differ by $2$, which means that if you know the ${n}^{\text{th}}$ terms, you will get the $n + {1}^{\text{th}}$ by subtracting two.

We start from ${a}_{0} = 7$, which is the starting point of the sequence. The next term, ${a}_{1}$, will be ${a}_{0} - 2 = 7 - 2 = 5$, and so on.

The general rule is what we just described with words: start from the initial value $7$, and subtract $2$ with each iteration. This means that, after $k$ iterations, we will have subtracted two $k$ times, i.e. we will have subtracted a total of $2 k$.

So, the rule is

${a}_{n} = 7 - 2 n$

You can confirm this by building some terms using the definition: given the starting value ${a}_{0}$, we have

${a}_{\setminus} \textcolor{red}{1} = {a}_{0} - \setminus \textcolor{red}{1} \cdot 2$
${a}_{\setminus} \textcolor{red}{2} = {a}_{1} - 2 = \left({a}_{0} - 2\right) - 2 = {a}_{0} - \setminus \textcolor{red}{2} \cdot 2$
${a}_{\setminus} \textcolor{red}{3} = {a}_{2} - 2 = \left({a}_{0} - 2 \cdot 2\right) - 2 = {a}_{0} - \setminus \textcolor{red}{3} \cdot 2$
${a}_{\setminus} \textcolor{red}{4} = {a}_{3} - 2 = \left({a}_{0} - 3 \cdot 2\right) - 2 = {a}_{0} - \setminus \textcolor{red}{4} \cdot 2$

as you can see, the index of the term is equal to the times we have to subtract $2$.