In a geometric sequence $t_{3} = 9$ $t_3=9$ and $t_{6} = \frac{9}{8}$ $t_6 =9/8$ how do you find $t_{12}$ $t_12$ ?

Question

1593 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-08T02:38:03

The general term for a geometric series is given by $t_{n} = a \times r^{n - 1}$ $t_n = a xx r^(n - 1)$

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

$9 = a \times r^{2}$ $9 = a xx r^(2)$

$\frac{9}{8} = a \times r^{5}$ $9/8 = a xx r^(5)$

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

$\frac{9}{r^{2}} = a$ $9/(r^2) = a$

$:. 9/(r^2) = 9/(8r^5)$

$0 = 9/(8r^5) - 9/(r^2)$

$0 = 9 - 72r^3$

$1/8 = r^3$

$r = 1/2$

$:. a = 9/(1/2)^2$

$a = 36$

$:. t_12 = a xx r^(n - 1)$

$t_12 = 36 xx 1/2^(12 - 1)$

$t_12 = 36 xx 1/2048$

$t_21 = 18/1024$

Note: The symbol $:.$ 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!

In a geometric sequence t_3=9t3=9t_3=9 and t_6 =9/8t6=98t_6 =9/8 how do you find t_12t12t_12?