Question

determine o 15° termo da pg (256,128,64...)​

Answer 1

Resposta:

Explicação passo a passo:

a1=256

q=128:256

q=1/2

a15=

an=a1.q^(n-1)

a15=256.(1/2)^(15-1)

a15=256.(1/2)^14

a15=256.1/16384

a15=256/16384 [ simplifica por 256]

a15=1/64

Answer 2

$> resolucao \\ \\ \geqslant progressao \: \: geometrica \\ \\ q = \frac{a2}{a1} \\ q = \frac{128}{256} \\ q = \frac{1}{2} \\ \\ > \: \: o \: 15 \: termo \: da \: pg \\ \\ \\ an = a1 \times q {}^{n - 1} \\ an = 256 \times ( \frac{1}{2} ) {}^{15 - 1} \\ an = 256 \times ( \frac{1}{2} ) {}^{14} \\ an = 256 \times \frac{1}{16384} \\ an = \frac{256}{16384} \\ an = \frac{1}{64} \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant \geqslant \geqslant$