Question

Na P.G infinita [24,12,6,...], determine
A) o 5• termo
B) a razão entre o 10• e o 8• termo nessa ordem;
C) o termo geral ou anesimo termo An.

Answer 1

a) $a_{5}=24* (\frac{1}{2} )^(5-1) \\ a_{5}=24* (\frac{1}{2} )^4\\ a_{5}=24* \frac{1}{16} \\ a_{5}= \frac{24}{16} \\ a_{5} = \frac{3}{2}$

b) $a_{10}=24* (\frac{1}{2})^(10-1) \\ a_{10}=24* (\frac{1}{2})^9 \\ a_{10}=24* \frac{1}{512} \\ a_{10}= \frac{24}{512} \\ a_{10}= (\frac{24}{512} )^{:8} \\a_{10}= \frac{3}{64}$

$a_{8}=24* (\frac{1}{2})^{10-1} \\ a_{8}=24* (\frac{1}{2})^{9} \\ a_{8}=24* (\frac{1}{128}) \\ a_{8}=\frac{24}{128}$
$a_{8}=24* (\frac{1}{2})^{10-1} \\ a_{8}=24* (\frac{1}{2})^{9} \\ a_{8}=24* (\frac{1}{128}) \\ a_{8}=(\frac{24}{128})^{:8} \\ a_{8}=\frac{3}{16}$

$\frac{a_{10}}{a_{8}} = \frac{ \frac{3}{64} }{ \frac{3}{16} } \\ \frac{a_{10}}{a_{8}} = \frac{3}{64} * \frac{16}{3}$

eliminamos o 3 por está no numerador e denominador, ficando 16/64

$\\ \frac{a_{10}}{a_{8}} = \frac{16}{64} = (\frac{16}{64})^{4}= \frac{1}{4}$

c) $a_{n}=24*( \frac{1}{2})^{n-1}$