Question

quantos termos tem a P.G de razão 4 cujos extremos são 5 e 1280?

Answer 1

$a_1=5 \\ a_n=1280 \\ q=4 \\ n=? \\ a_n=a_1.q^{(n-1)} \\ 1280=5.4^{(n-1)} \\ \\ \frac{1280}{5}=4^{(n-1)} \\ \\ 4^4=4^{(n-1)} \\ \\ 4=n-1 \\ \\ 4+1=n=5 \\ \\ n=5$

PROVA REAL

$a_n=5.4^{(5-1)} \\ \\ a_n=5.4^4 \\ \\ a_n=5.256 \\ \\ a_n=1280$