Question

Determinar o número de termos da P.G. (1000, - 500, ... , -125/16) : *

8
10
6
5

Answer 1

$> resolucao \\ \\ \geqslant progressao \\ \\ q = \frac{a2}{a1} \\ \\ q = \frac{ - 500}{1000} \\ \\ q = \frac{ - 1}{2} \\ \\ \\ an = a1 \times q {}^{n - 1} \\ \frac{ - 125}{16} = 1000 \times ( - \frac{1}{2} ) {}^{n - 1} \\ \frac{ \frac{ - 125}{16} }{1000} = ( - \frac{1}{2} ) {}^{n - 1} \\ - \frac{125}{16} \times \frac{1}{1000} = ( - \frac{1}{2} ) {}^{n - 1} \\ - \frac{125}{16000} = -( \frac{1}{2} ) {}^{n - 1} \\ - \frac{1}{128} = - ( \frac{1}{2} ) {}^{n - 1} \\ - ( \frac{1}{2}) {}^{7} = - ( \frac{1}{2} ) {}^{n - 1} \\ n - 1 = 7 \\ n = 7 + 1 \\ n = 8 \\ \\ \\ resposta \: > \: pg \: de \: 8 \: termos \\ \\ > < > < > < > < > < >$