Question

calcule a razao de uma PG de seis termos cujos extremos são 3 e 96

Answer 1

$n=6\\a_1=3\\a_n=96\\q=?\\\\\\ a_n=a_1*q^{n-1}\\\\\\ 96=3*q^{6-1}\to \\\\ 96=3*q^{5}\to \\\\ \frac{96}{3} =q^5\to \\\\ 32=q^5\to \\\\ 2^5=q^5\to \\\\ q=2\\\\\\\\ PG(3, 6, 12, 24, 48, 96)$

Answer 2

a₁ = 3
a₆ = 96
n = 6
q = ?
      a₆ = a₁ * q^n-1
      96 = 3 * q⁶⁻¹
      96 = 3 * q⁵
      96 ÷ 3 = q⁵
      32 = q⁵
      q = $\sqrt[5]{32}$
              32 | 2
              16 | 2
                8 | 2
                4 | 2
                2 | 2                                       
                1 | $\sqrt[5]{2^5} = 2$
      q = 2 

Espero ter ajudado. Valeu!