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Sagot :
Answer:
6n-5
Step-by-step explanation:
Some different types of sequences
To find sequence patterns, often the sequences are arithmetic (adding by a constant amount) or geometric (multiplying by a constant amount).
To test if something is a geometric sequence, find the quotient of (divide) adjacent terms.
To test if something is an arithmetic sequence, find the difference of (subtract) adjacent terms.
Finding the type of our sequence
Note the following:
19-13=6
13-7=6
7-1=6
So, to get each next term, add 6. However, what is the nth term?
Finding an expression for our sequence
You'll be adding up "n" sixes to get there (expressed with multiplication, because multiplication is shorthand for repeated addition), but we need a starting point so that when n=1, the expression equals 1, and when n=2, the expression equals 7... and so on. This starting point shifts the "6n" piece of our expression so that it lines up on the sequence. I'm going to call this starting point "b" for the base that is supporting this sequence.
So the expression will look something like [tex]6n+b[/tex]
Knowing that the expression needs to equal 1 when n=1, we can set up an equation and solve for "b".
[tex]6n+b=n^{\text{th}} \text{ term}\\6(1)+b=(1)\\6+b=1\\(6+b)-6=(1)-6\\b=-5[/tex]
So, the expression for the nth term of the sequence is [tex]6n-5[/tex]
Verifying
To verify that the expression we found is right, we can test our expression for each of the terms we do know:
The first term (n=1) is 1
[tex]6n+b=n^{\text{th}} \text{ term}\\[/tex]
[tex]6(1)-5 \overset{?}{=} (1)[/tex]
[tex]6-5 \overset{?}{=} 1[/tex]
[tex]1 \overset{\checkmark}{=} 1[/tex]
The second term (n=2) is 7
[tex]6n+b=n^{\text{th}} \text{ term}\\[/tex]
[tex]6(2)-5 \overset{?}{=} (7)[/tex]
[tex]12-5 \overset{?}{=} 7[/tex]
[tex]7 \overset{\checkmark}{=} 7[/tex]
The third term (n=3) is 13
[tex]6n+b=n^{\text{th}} \text{ term}\\[/tex]
[tex]6(3)-5 \overset{?}{=} (13)[/tex]
[tex]18-5 \overset{?}{=} 13[/tex]
[tex]13 \overset{\checkmark}{=} 13[/tex]
The fourth term (n=4) is 19
[tex]6n+b=n^{\text{th}} \text{ term}\\[/tex]
[tex]6(4)-5 \overset{?}{=} (19)[/tex]
[tex]24-5 \overset{?}{=} 19[/tex]
[tex]19 \overset{\checkmark}{=} 19[/tex]
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