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