using our definitions:
x/3 = y/x ----> x^2 = 3y
and
y/x = 18/y ----> y^2 = 18x
so we get x/3 = 18/y
xy = 54 -----> y = 54/x
in x^2 = 3y
x^2 = 3(54/x)
x^3 = 162 = 6(27)
x =3(6^(1/3)) or appr 5.45136 ------> 3 times the cube root of 6
back in y = 54/x
= 54/[3(6^(1/3))]
= 18/6^(1/3)
= appr 9.90578
Given that 3, x, y, 18 are in a geometric progression. find the values of X and Y.
2 answers
The terms n-th term of a geometric sequence:
an = a1 ∙ rⁿ⁻¹
where:
a1 = first term
r = common ratio
In this case:
a1 = 3
a2 = x
a3 = y
a4 = 18
a4 = a1 ∙ r ⁴⁻¹ = a1 ∙ r³
18 = 3 ∙ r³
r³ = 18 / 3 = 6
r = ∛6
a2 = a1 ∙ r
x = 3 ∙ r
x = 3 ∙ ∛6
a3 = a1 ∙ r²
y = 3 ∙ ( ∛6 )²
y = 3 ∙ 6²/³
a4 = a1 ∙ r³
a4 = 3 ∙ ( ∛6 )³
a4 = 3 ∙ 6 = 18
So:
x = 3 ∙ ∛6 , y = 3 ∙ 6²/³
an = a1 ∙ rⁿ⁻¹
where:
a1 = first term
r = common ratio
In this case:
a1 = 3
a2 = x
a3 = y
a4 = 18
a4 = a1 ∙ r ⁴⁻¹ = a1 ∙ r³
18 = 3 ∙ r³
r³ = 18 / 3 = 6
r = ∛6
a2 = a1 ∙ r
x = 3 ∙ r
x = 3 ∙ ∛6
a3 = a1 ∙ r²
y = 3 ∙ ( ∛6 )²
y = 3 ∙ 6²/³
a4 = a1 ∙ r³
a4 = 3 ∙ ( ∛6 )³
a4 = 3 ∙ 6 = 18
So:
x = 3 ∙ ∛6 , y = 3 ∙ 6²/³