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

Question

1447 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 xx r^(n - 1)#

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

#9 = a xx r^(2)#

#9/8 = a xx r^(5)#

#9/(8r^5) = 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!