How do you write a rule for the nth term of the geometric sequence, then find a6 given 2, 4/3, 8/9, 16/27,...?

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How do you write a rule for the nth term of the geometric sequence, then find a6 given 2, 4/3, 8/9, 16/27,...?

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Answer 1 · 2016-05-31T21:28:05

$T_{6} = \frac{64}{243}$ $T_6 = 64/243$

Explanation:

The first term is given as 2, so $a = 2$ $a = 2$

The common ration, $r$ $r$ is found by dividing 2 consecutive terms..
$r = \frac{T_{n}}{T_{n - 1}}$ $r = (T_n)/(T_(n-1))$ In other words, a term divided by the one before it.

$r = \frac{4}{3} \div \frac{2}{1} = \frac{4}{3} \times \frac{1}{2} = \frac{2}{3}$ $r = 4/3 ÷2/1 = 4/3 xx 1/2 = 2/3$

Check: $r = \frac{8}{9} \div \frac{4}{3} = \frac{8}{9} \times \frac{3}{4} = \frac{2}{3}$ $r = 8/9 ÷ 4/3 = 8/9 xx 3/4 =2/3$

Now, knowing the values for $a and r$ $a and r$ , we can substitute into the general form for a term of a GP

$T_{n} = a r^{n - 1} \Rightarrow 2 \times {(\frac{2}{3})}^{n - 1}$ $T_n = ar^(n-1) rArr 2 xx (2/3)^(n-1)$

Simplifying gives: $T_{n} = \frac{2^{n}}{3^{n - 1}}$ $T_n =(2^n)/(3^(n-1)$

In the same way we can find $a_{6} = a r^{5}$ $a_6 = ar^5$

But we have already found the general form, so let's use that for $n = 6$ $n = 6$

$T_{6} = \frac{2^{6}}{3^{5}}$ $T_6 = (2^6)/(3^5)$

$T_{6} = \frac{64}{243}$ $T_6 = 64/243$

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How do you write a rule for the nth term of the geometric sequence, then find a6 given 2, 4/3, 8/9, 16/27,...?

1 Answer

Explanation:

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