4,x,y,32 are in GP, find x,y and the 11th term of the progression.

Answer 1

r = x/4
r = y/x
r = 32/y

x/4 = y/x
x^2 = 4y ----> y = x^2/4

32/y = y/x
y^2 = 32x

then : x^4/16 = 32x
x^3 = 512
x = 8
y = 64/4 = 16

the sequence is 4,8, 16, 32, ...
term11 = ar^10 = 4(2)^10 = 4096

OR, even simpler ....

a = 4
ar^4 = 32
4r^3 = 32
r^3 = 8
r = 2
so the sequence is 4, 8, 16, 32, ...
x = 8, y = 16
etc