Find the values of k for which 6,x,y 162 are geometric progression.

Answer 1

The n-th term of a GP

where:

a = a1 initial value

r = common ratio

an = a r ⁿ ⁻ ¹

In this case a1 = a = 6 , a4 = 162

so

a4 = 6 ∙ r ⁴ ⁻ ¹

162 = 6 ∙162 = 6 ∙ r ³

162 / 6 = r ³

27 = r ³

r = ∛ 27

r = 3

x = a2 = a ∙ r = 6 ∙ 3 = 18

y = a3 = a ∙ r ² = 6 ∙ 3 ² = 6 ∙ 9 = 54

Your GP:

6 , 18 , 54 , 162