Question

Qual a razão de uma progressão geométrica em que a1 = 5 a4 = 135?

Answer 1

$a_1=5\\ a_4=135\\ n=4\\ q=?\\\\\\ a_n=a_1\times q^{n-1}\\ \\\\ 135=5\times q^{4-1}\to\\\\ 135=5\times q^3\to\\\\ \dfrac{135}{5}=q^3\to \\\\ 27=q^3\to\\\\ 3^{\not 3}=q^{\not 3}\to\\\\ \large\boxed{q=3} \\\\\\\\\\ PG~\{ 5, 15, 45, 135 \}$