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

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

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Answer 1 · 2018-06-27T08:49:32

each term is going down by 2

$\Rightarrow - 2 n$ $=> -2n$

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

to get the sequence we need to find the adjuster

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

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

Answer 2 · 2018-06-27T08:49:50

$a_{n} = 7 - 2 n$ $a_n=7-2n$

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 $2$ $2$ , which means that if you know the $n^{th}$ $n^"th"$ terms, you will get the $n + 1^{th}$ $n+1^"th"$ by subtracting two.

We start from $a_{0} = 7$ $a_0=7$ , which is the starting point of the sequence. The next term, $a_{1}$ $a_1$ , will be $a_{0} - 2 = 7 - 2 = 5$ $a_0-2=7-2=5$ , and so on.

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

So, the rule is

$a_{n} = 7 - 2 n$ $a_n=7-2n$

You can confirm this by building some terms using the definition: given the starting value $a_{0}$ $a_0$ , we have

$a_{1} = a_{0} - 1 \cdot 2$ $a_\color(red)(1) = a_0-\color(red)(1)*2$
$a_{2} = a_{1} - 2 = (a_{0} - 2) - 2 = a_{0} - 2 \cdot 2$ $a_\color(red)(2) = a_1-2 = (a_0-2)-2 = a_0-\color(red)(2)*2$
$a_{3} = a_{2} - 2 = (a_{0} - 2 \cdot 2) - 2 = a_{0} - 3 \cdot 2$ $a_\color(red)(3) = a_2-2 = (a_0-2*2)-2 = a_0-\color(red)(3)*2$
$a_{4} = a_{3} - 2 = (a_{0} - 3 \cdot 2) - 2 = a_{0} - 4 \cdot 2$ $a_\color(red)(4) = a_3-2 = (a_0-3*2)-2 = a_0-\color(red)(4)*2$

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

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

2 Answers

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