How do I find the sixth term of a geometric sequence for which #t_5 = 24# and #t_8 = 3#?

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Answer 1 · 2015-10-31T16:16:32

I have shown how to arrive at the needed information but left some of the working out for you to do.

#t_8<t_5# so Thus reducing. That means that the ratio is less than one.

Let k be a constant
Let the ratio be #1/r#

giving:
#t_5 = k(1/r)^5 = 24# ........................( 1 )
#t_8=k(1/r)^8 = 3# .............................( 2 )

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To find #1/r# divide equation 1 by equation 2 giving

#(1/r)^(5-8)=(24)/(3) =8#

#(1/r)^(-3) =8#

#r^3=8#
#r=2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To find k
Substitute for r in equation ( 1 ) giving

#k(1/2)^5=24#

#k= (2^5)(24)# "I will let you work that out!"
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We can now answer the question:

#t_6= (2^5)(24)(1/2)^6# " Again, I will let you work that out."

By the way: #2^5 times 1/(2^6) = (2^5)/(2^6) = 1/2#
So #1/2 times 12#