The first three terms of the sequence 6;x;y;27 form an arithmetic progression and the last three terms form a geometric progression. Determine the values of x and y
3 answers
x = 15 and y = 18
x = 6 + a
y = 6 + 2 a
y = x r
27 = x r^2
x r = 6 + 2 a
so
(6+a) r = 6 + 2a so r = (6+2a) / (6+a)
and
27 = (6+a) r^2
27 = (6+a) (6+2a)^2 / (6+a)^2 = (6^2 + 24 a + 4a^2)/(6+a)
162 + 27 a = 4 a^2 + 24 a + 36
4 a^2 -3 a -126 = 0
positive solution of quadratic for a = 6
so
x = 12
y = 18
y = 6 + 2 a
y = x r
27 = x r^2
x r = 6 + 2 a
so
(6+a) r = 6 + 2a so r = (6+2a) / (6+a)
and
27 = (6+a) r^2
27 = (6+a) (6+2a)^2 / (6+a)^2 = (6^2 + 24 a + 4a^2)/(6+a)
162 + 27 a = 4 a^2 + 24 a + 36
4 a^2 -3 a -126 = 0
positive solution of quadratic for a = 6
so
x = 12
y = 18
or
y - x = x - 6
y = 2x - 6
y^2 = 27x
(2x-6)^2 = 27x
4x^2 - 24x + 36 = 27x
4x^2 - 51x + 36 = 0
(x - 12)(4x - 3) = 0
x = 12 or x = 3/4
if x =12, y = 18
if x = 3/4 , y = -9/2
y - x = x - 6
y = 2x - 6
y^2 = 27x
(2x-6)^2 = 27x
4x^2 - 24x + 36 = 27x
4x^2 - 51x + 36 = 0
(x - 12)(4x - 3) = 0
x = 12 or x = 3/4
if x =12, y = 18
if x = 3/4 , y = -9/2