Question

determine o 9° termo da pg. (405,135,45...)​

Answer 1

Resposta:

P.G(405,135,45...)​

Calcular a razão

$a_1=405\\ \\ a_2=135\\ \\ q=\dfrac{135}{405}\\ \\ simplificando\\ \\\boxed{ q=\dfrac{1}{3}}$

Podemos notar que a P.G está decrescente

Sendo:

$a_n=a_9=?\\ a_1=405\\ n=9\\ \\ q=\dfrac{1}{3}\\ \\ \\ Termo~~geral\\ \\ a_n=a_1.q^{n-1}\\ \\ \\ a_9=405.(\dfrac{1}{3})^{9-1}\\ \\ \\ a_9=405.(\dfrac{1}{3})^8\\ \\ \\ a_9=3^4.5.\dfrac{1}{3^8}\\ \\ a_9=\dfrac{5}{3^4}\\ \\ \\\boxed{ a_9=\dfrac{5}{81}}$

Answer 2

$> \: resolucao \\ \\ \geqslant \: progressao \: \: geometrica \\ \\ q = \frac{a2}{a1} = \frac{135}{405} = \frac{1}{3} \\ \\ = = = = = = = = = = = = = = = = \\ \\ > \: o \: nono \: termo \: da \: pg \\ \\ an = a1 \times q {}^{n - 1} \\ an = 405 \times ( \frac{1}{3} ) {}^{9 - 1} \\ an = 405 \times ( \frac{1}{3} ) {}^{8} \\ an = 405 \times \frac{1}{6561} \\ \\ an = \frac{405÷81}{6561÷81} \\ \\ an = \frac{5}{81} \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant$