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

Oct 31, 2015

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 $\frac{1}{r}$

giving:
${t}_{5} = k {\left(\frac{1}{r}\right)}^{5} = 24$ ........................( 1 )
${t}_{8} = k {\left(\frac{1}{r}\right)}^{8} = 3$ .............................( 2 )

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To find $\frac{1}{r}$ divide equation 1 by equation 2 giving

${\left(\frac{1}{r}\right)}^{5 - 8} = \frac{24}{3} = 8$

${\left(\frac{1}{r}\right)}^{- 3} = 8$

${r}^{3} = 8$
$r = 2$
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To find k
Substitute for r in equation ( 1 ) giving

$k {\left(\frac{1}{2}\right)}^{5} = 24$

$k = \left({2}^{5}\right) \left(24\right)$ "I will let you work that out!"
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We can now answer the question:

${t}_{6} = \left({2}^{5}\right) \left(24\right) {\left(\frac{1}{2}\right)}^{6}$ " Again, I will let you work that out."

By the way: ${2}^{5} \times \frac{1}{{2}^{6}} = \frac{{2}^{5}}{{2}^{6}} = \frac{1}{2}$
So $\frac{1}{2} \times 12$