Respuesta :
Answer:
Series First term(a_1) common ratio(r)
1,3,9,27,... 1 3
8,4,2,1,.... 8 1/2
4,-16,64,-256,... 4 -4
Step-by-step explanation:
We know that the nth term of a geometric series is given by:
[tex]a_n=a_1\cdot r^{n-1}[/tex]
where [tex]a_1[/tex] is the first term and r is the common ratio.
1)
The first sequence is:
1,3,9,27,....
The first term i.e. [tex]a_1=1[/tex]
and if r is the common ratio then,
[tex]a_2=a_1\cdot r\\\\i.e.\\\\r=\dfrac{a_2}{a_1}[/tex]
Since, from the sequence we have:
[tex]a_2=3[/tex]
This means that:
[tex]r=\dfrac{3}{1}\\\\i.e.\\\\r=3[/tex]
2)
Second sequence is:
8,4,2,1,....
The first term i.e. [tex]a_1=8[/tex]
and if r is the common ratio then,
[tex]a_2=a_1\cdot r\\\\i.e.\\\\r=\dfrac{a_2}{a_1}[/tex]
Since, from the sequence we have:
[tex]a_2=4[/tex]
This means that:
[tex]r=\dfrac{4}{8}\\\\i.e.\\\\r=\dfrac{1}{2}[/tex]
3)
Third sequence is:
4,-16,64,-256,...
The first term i.e. [tex]a_1=4[/tex]
and if r is the common ratio then,
[tex]a_2=a_1\cdot r\\\\i.e.\\\\r=\dfrac{a_2}{a_1}[/tex]
Since, from the sequence we have:
[tex]a_2=-16[/tex]
This means that:
[tex]r=\dfrac{-16}{4}\\\\i.e.\\\\r=-4[/tex]
The nth term of a geometric sequence is [tex]a_n = a_1 \times r^{ n-1}[/tex] where a1 is the first term and r is the common ratio.
Sequence 1 = 1, 3, 9, 27, .......
For the first sequence, a1 = 1, r = 3.
Sequence 2 = 8, 4, 2, 1, .......
For the second sequence, a1 = 8, r = 1/2.
Sequence 3 = 4, -16, 64, 256, .......
For the third sequence, a1 = 4, r = -4.
What is a geometric sequence?
A geometric sequence can be defined as a sequence that has a constant ratio between each pair of numbers.
Given that the nth term of a geometric sequence is an = a1 • r^ n-1, where a1 is the first term and r is the common ratio.
Sequence 1:
The first geometric sequence is 1, 3, 9, 27, .......
We know that the nth term is given as [tex]a_n = a_1\times r^{n-1}[/tex].
Here a_1 is the first term and r is the common ratio. The second term is given as,
[tex]a_2 = a_1\times r^{2-1}[/tex]
[tex]a_2 = a_1 r[/tex]
So for the given sequence, a1 = 1, a2 = 3, then the common ratio is given as,
[tex]3 = 1\times r^{2-1}[/tex]
[tex]r = 3[/tex]
Hence for the first sequence, a1 = 1, r = 3.
Sequence 2:
The second geometric sequence is 8, 4, 2, 1, .......
For given sequence, a1= 8, a2 = 4, then the common ratio is,
[tex]4 = 8\times r^{2-1}[/tex]
[tex]r = \dfrac {4}{8}[/tex]
[tex]r = \dfrac{1}{2}[/tex]
Hence for the second sequence, a1 = 8, r = 1/2.
Sequence 3:
The third geometric sequence is 4, -16, 64, 256, .......
For given sequence, a1= 4, a2 = -16, then the common ratio is,
[tex]-16 = 4\times r^{2-1}[/tex]
[tex]r = \dfrac {-16}{4}[/tex]
[tex]r = -4[/tex]
Hence for the third sequence, a1 = 4, r = -4.
To know more about the geometric sequence, follow the link given below.
https://brainly.com/question/4069689.
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