SOLVE FOR THE COMMON RATIO
Solve for the common ratio of the sequence where 7th term is 6 and the 11th term is [tex] \frac{3}{8} [/tex]
FORMULA:
[tex]r = \sqrt[4]{ \frac{b}{a} } [/tex]
SOLUTION:
[tex]r = \sqrt[4]{ \frac{b}{a} } \\ \\ r \: = \sqrt[4]{ \frac{ \frac{3}{8} }{6} } \\ \\ r = \sqrt[4]{ \frac{1}{16} } \\ \\ r = \frac{1}{2} [/tex]
- The common ratio is [tex] \frac{1}{2} [/tex].
To check, just multiply the common ratio by the given first extreme.
6 - 7th term
[tex]6 \times \frac{1}{2} = \frac{6}{2} \div 2 = 3[/tex] - 8th term
[tex]3 \times \frac{1}{2} = \frac{3}{2} [/tex] - 9th term
[tex] \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} [/tex] - 10th term
[tex] \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} [/tex] - 11th term ✔︎
ANSWER:
The common ratio is [tex] \frac{1}{2} [/tex]
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