Question

What is the common ratio of the geometric sequence whose second and fourth terms are 6 and 54, respectively?

Answer 1

a₂ = 6 a₄ = 54


`a_(n) = q^(n-1) * a_(1)`


`\left \{ {{ a_(2) = q^(2-1) * a_(1) } \atop { a_(4) = q^(4-1)* a_(1) }} \right. \\ \\ \left \{ {{6 = q * a_(1) } \atop {54 = q^(3) * a_(1) }} \right. \\ \\ \left \{ {{ a_(1) = (6)/(q) } \atop {54 = q^(3) * (6)/(q) }} \right. \\ \\ \left \{ {{ a_(1) = (6)/(q) } \atop {54 = q^(2) * 6 }} \right.`


`\left \{ {{ a_(1) = (6)/(q) } \atop { q^(2) =9}} \right. \\ \\ \left \{ {{ a_(1) = (6)/(q) } \atop {q= √(9) }} \right.`
q = 3 q = -3

a₁ = 6/3 = 2 a₁ = 6/-3 = -2

Answer 2

Hi there! T4=T2×r²,6r²=54. Therefore, the answer would be 3.