Question

Quantos termos tem a P.G. (2, 6, ...,1458)​

Answer 1

$> resolucao \\ \\ \geqslant progressao \: \: geometrica \\ \\ q = \frac{a2}{a1} \\ q = \frac{6}{2} \\ q = 3 \\ \\ \\ \\ an = a1 \times q {}^{n - 1} \\ 1458 = 2 \times 3 {}^{n - 1} \\ \frac{1458}{2} = 3 {}^{n - 1} \\ 729 = 3 {}^{n - 1} \\ 3 {}^{6} = 3 {}^{n - 1} \\ n - 1 = 6 \\ n = 6 + 1 \\ n = 7 \\ \\ \\ resposta \: > \: \: pg \: \: de \: \: 7 \: \: termos \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant \geqslant \geqslant$