Answer :
Answer:
[tex]\frac{1}{3}[/tex]
Step-by-step explanation:
Given:
The second term in the geometric sequence is 12.
The fourth term inn the same sequence is 4/3.
Question asked:
Common ratio = ?
Solution:
Let 1st term = a and common ratio = r
By using formula, [tex]a_{n} = ar^{n-1}[/tex]
For 2nd term, [tex]12 = ar ( equation 1 )[/tex]
Similarly, for the 4th term,[tex]\frac{4}{3} = ar^{3}[/tex] [tex]( equation 2 )[/tex]
Dividing equation 1 and 2
[tex]12\div\frac{4}{3}[/tex] = [tex]\frac{ar}{ar^{3} } \\[/tex]
[tex]12\times\frac{3}{4} = \frac{1}{r^{2} } \\\frac{36}{4} = \frac{1}{r^{2} }[/tex]
[tex]9 = \frac{1}{r^{2} }\\[/tex]
By cross multiplication
[tex]r^{2} = \frac{1}{9}[/tex]
By taking root both side
[tex]r = \sqrt[2]{\frac{1}{9} } \\r = \frac{1}{3}[/tex]
Thus, common ratio is [tex]\frac{1}{3}[/tex]