The solution to the geometric sequence in the question are;
First part;
r = 3
[tex]a_{8} = 8748 [/tex]
Second part;
r = 4
[tex]a_{6} = 9216 [/tex]
Practice:
r = 8
[tex]a_{7} = 0.3125 [/tex]
Third part;
Please see the detailed reasons in following section
Fourth part;
[tex] {a}_{n} = 4 × {a}_{n-1}[/tex]
[tex] {a}_{n} = 0.2 × {a}_{n-1}[/tex]
First part;
The formula for geometric progression can be presented as follows;
[tex] a_{n} = a_{1} \times {r}^{n - 1} [/tex]
[tex]a_{4} =108 = a_{1} \times {r}^{4 - 1} [/tex]
Where;
[tex]a_{1} = 4 \: and \: a_{4} = 108[/tex]
We have;
[tex]a_{4} =108 = 4 \times {r}^{4 - 1} = 4 \cdot {r}^{3} [/tex]
Which gives;
[tex] {r}^{3} = \frac{108}{4} = 27[/tex]
[tex]r = \sqrt[3]{27} = 3[/tex]
r = 3
Therefore;
[tex]a_{8} = 4 \times {3}^{7} = 8748 [/tex]
Second part;
Where;
[tex]a_{1} = 9 \: and \: a_{3} = 144[/tex]
Therefore;
[tex] {r}^{2} = \frac{144}{9} = 16[/tex]
[tex]r = \sqrt{16} = 4[/tex]
r = 4
Which gives;
[tex]a_{6} = 9 \times {4}^{5} = 9216 [/tex]
Practice;
[tex]a_{1} = 3 \: and \: a_{4} = 1536[/tex]
Therefore;
[tex] {r}^{3} = \frac{1536}{3} = 512[/tex]
[tex]r = \sqrt[3]{512} = 8[/tex]
r = 8
[tex]a_{1} = 20 \: and \: a_{3} = 5[/tex]
Therefore;
[tex] {r}^{2} = \frac{5}{20} = 0.25[/tex]
[tex]r = \sqrt{0.25} = 0.5[/tex]
Which gives;
[tex]a_{7} = 20 \times {0.5}^{6} = 0.3125 [/tex]
Third part
Recursively means expression in terms of previous terms
The meaning of the equation is that each term of a geometric sequence is given by the product of the common ratio and the previous term
Fourth part;
The recursive formula for the sequence, {2, 8, 32, 128...} is found as follows;
r = 8/2 = 4
Therefore;
[tex] {a}_{n} = 4 × {a}_{n-1}[/tex]
Part;
The sequence is {200, 40, 8, 1.6...}
The common ratio is; 40/200 = 0.2
The recursive formula is therefore;
[tex] {a}_{n} = 0.2 × {a}_{n-1}[/tex]
Learn more about geometric sequence here:
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