# In a geometric sequence t_3=9 and t_6 =9/8 how do you find t_12?

May 8, 2016

The general term for a geometric series is given by ${t}_{n} = a \times {r}^{n - 1}$

Thus, we write a systems of equations, and solve for r and a.

$9 = a \times {r}^{2}$

$\frac{9}{8} = a \times {r}^{5}$

$\frac{9}{8 {r}^{5}} = a$

$\frac{9}{{r}^{2}} = a$

$\therefore \frac{9}{{r}^{2}} = \frac{9}{8 {r}^{5}}$

$0 = \frac{9}{8 {r}^{5}} - \frac{9}{{r}^{2}}$

$0 = 9 - 72 {r}^{3}$

$\frac{1}{8} = {r}^{3}$

$r = \frac{1}{2}$

$\therefore a = \frac{9}{\frac{1}{2}} ^ 2$

$a = 36$

$\therefore {t}_{12} = a \times {r}^{n - 1}$

${t}_{12} = 36 \times \frac{1}{2} ^ \left(12 - 1\right)$

${t}_{12} = 36 \times \frac{1}{2048}$

${t}_{21} = \frac{18}{1024}$

Note: The symbol $\therefore$ denotes a change of idea, just as a new paragraph in text is meant to demonstrate the introduction of a new thought.

Hopefully this helps!