Question

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

Answer 1

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

Explanation:

`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 )

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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`
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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!"
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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`