The 3rd and 6th terms of a geometric progression (G.P.) are \(\frac{8}{3}\) and \(\frac{64}{81}\) respectively, find the common ratio.
- \(\frac{1}{3}\)
-
\(\frac{2}{3}\)
- \(\frac{3}{4}\)
- \(\frac{4}{3}\)
The correct answer is: B
Explanation
\(T_{n} = ar^{n-1}\) (for a geometric progression)
\(T_{3} = ar^{3-1} = ar^{2} = \frac{8}{3}\)
\(T_{6} = ar^{6-1} = ar^{5} = \frac{64}{81}\)
Dividing \(T_{6}\) by \(T_{3}\),
\(\frac{ar^{5}}{ar^{2}} = \frac{\frac{64}{81}}{\frac{8}{3}} \implies r^{3} = \frac{8}{27}\)
\(\therefore r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)