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

Answers

Answer 1

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

Answer 2

Answer: a₂ = 6 a₄ = 54

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

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

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

a₁ = 6/3 = 2 a₁ = 6/-3 = -2

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Answers

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Answers

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Answers

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