Question

13. Na P.G. (3, 6, 12, ...) calcule:

a) o décimo termo;

b) a ordem do termo de valor seis mil cento e quarenta e quatro.​

Answer 1

$> resolucao \\ \\ \geqslant progressao \: \: geometrica \\ \\ q = \frac{a2}{a1} \\ q = \frac{6}{3} \\ q = 2 \\ \\ \\ > o \: decimo \: termo \: da \: pg \\ \\ an = a1 \times q {}^{n - 1} \\ an = 3 \times 2 {}^{10 - 1} \\ an = 3 \times 2 {}^{9} \\ an = 3 \times 512 \\ an = 1536 \\ \\ \\ = = = = = = = = = = = = = = = = = = = = = = \\ \\ \\ > ordem \: do \: termo \: de \: valor \: 6144 \\ \\ \\ an = a1 \times q {}^{n - 1} \\ 6144 = 3 \times 2 {}^{n - 1} \\ \frac{6144}{3} = 2 {}^{n - 1} \\ 2048 = 2 {}^{n - 1} \\ 2 {}^{11} = 2 {}^{n - 1} \\ n - 1 = 11 \\ n = 11 + 1 \\ n = 12 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant \geqslant \geqslant$