Answer:
The common ratio is [tex]\frac{\sqrt{3} }{3}[/tex] and the next three terms of the sequence are [tex]\sqrt{3}, 1, \frac{1}{\sqrt{3} }[/tex]
Step-by-step explanation:
Given the geometric series a1, a2, a3... the common ratio is expressed as;
r = a2/a1 = a3/a2
The nth term of a geometric sequence Tn = [tex]ar^{n-1}[/tex]
where n is the number of terms
r is the common ratio
Now given the sequence 9,3[tex]\sqrt{3}[/tex], 3...
common ratio [tex]r=\frac{3\sqrt{3} }{9} = \frac{3}{3\sqrt{3} } = \frac{\sqrt{3} }{3}[/tex]
The next three terms are the 4th, 5th and 6th term
To get the 4th term when n = 4
[tex]=9*(\frac{\sqrt{3} }{3}) ^{4-1}\\=9*(\frac{\sqrt{3} }{3}) ^{3}\\\\= 9*\frac{\sqrt{27} }{27}\\= \frac{3\sqrt{3} }{3}\\ T4 = \sqrt{3} \\[/tex]
When n= 5
[tex]T5 =9*(\frac{\sqrt{3} }{3}) ^{5-1}\\T5=9*(\frac{\sqrt{3} }{3}) ^{4}\\= 9*\frac{\sqrt{81} }{81}\\ T5= \frac{81}{81} \\T5 = 1[/tex]
when n = 6
[tex]=9*(\frac{\sqrt{3} }{3}) ^{6-1}\\=9*(\frac{\sqrt{3} }{3}) ^{5}\\\\= 9*\frac{\sqrt{243} }{243}\\ = 9*\frac{9\sqrt{3} }{243} \\= \frac{81\sqrt{3} }{243} \\= \frac{\sqrt{3} }{3} \\T6 = \frac{1}{\sqrt{3} }[/tex]
The next three terms of the sequence are [tex]\sqrt{3}, 1, \frac{1}{\sqrt{3} }[/tex]
