Question

dada a PG, onde a10= 1024 e a razão é 2, calcule a1​

Answer 1

$> resolucao \\ \\ \geqslant progressao \: \: geometrica \\ \\ an = a1 \times q {}^{n - 1} \\ 1024 = a1 \times 2 {}^{10 - 1} \\ 1024 = a1 \times 2 {}^{9} \\ 1024 = a1 \times 512 \\ a1 = \frac{1024}{512} \\ a1 = 2 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant \geqslant$