How do you write the expression for the nth term of the geometric sequence $a_{3} = \frac{16}{3}, a_{5} = \frac{64}{27}, n = 7$ $a_3=16/3, a_5=64/27, n=7$ ?

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How do you write the expression for the nth term of the geometric sequence $a_{3} = \frac{16}{3}, a_{5} = \frac{64}{27}, n = 7$ $a_3=16/3, a_5=64/27, n=7$ ?

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Answer 1 · 2018-05-14T00:39:50

$a_{n} = \frac{2^{n + 1}}{3^{n - 2}}$ $a_n=2^(n+1)/3^(n-2)$
So:
$a_{7} = \frac{256}{243}$ $a_7=256/243$

Explanation:

$a_{3} = \frac{16}{3}, a_{5} = \frac{64}{27}$ $a_3=16/3,a_5=64/27$

$a_{5} = \frac{16 \cdot 2 \cdot 2}{3 \cdot 3 \cdot 3} = a_{3} \cdot \frac{2}{3} \cdot \frac{2}{3}$ $a_5=( 16 * 2 * 2 )/( 3 * 3 * 3 )=a_3 *2/3*2/3$

So: $a_{n + 1} = a_{n} \cdot \frac{2}{3}$ $a_(n+1)=a_n * 2/3$

That will mean:
$a_{3} = a_{2} \cdot \frac{2}{3} \Rightarrow a_{2} = \frac{a_{3}}{\frac{2}{3}} = \frac{\frac{16}{2}}{\frac{3}{3}} \Rightarrow a_{2} = \frac{8}{1}$ $a_3=a_2*2/3 rArr a_2=a_3/ (2/3)=(16/2) / (3/3) rArr a_2=8/1$

So, we have:
$a_{2} = \frac{2^{3}}{3^{0}}, a_{3} = \frac{2^{4}}{3^{1}}, a_{5} = \frac{2^{6}}{3^{3}}$ $a_2=2^3/3^0, a_3=2^4/3^1, a_5=2^6/3^3$

We can deduct:
$a_{n} = \frac{2^{n + 1}}{3^{n - 2}}$ $a_n=2^(n+1)/3^(n-2)$

So:
$a_{7} = \frac{2^{8}}{3^{5}} = \frac{256}{243}$ $a_7=2^8/3^5=256/243$

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How do you write the expression for the nth term of the geometric sequence $a_{3} = \frac{16}{3}, a_{5} = \frac{64}{27}, n = 7$ $a_3=16/3, a_5=64/27, n=7$ ?

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How do you write the expression for the nth term of the geometric sequence a_3=16/3, a_5=64/27, n=7a3=163,a5=6427,n=7a_3=16/3, a_5=64/27, n=7?

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How do you write the expression for the nth term of the geometric sequence $a_{3} = \frac{16}{3}, a_{5} = \frac{64}{27}, n = 7$ $a_3=16/3, a_5=64/27, n=7$ ?