It is given that x, 2y, 3z are in A.P
∴ 2y – x = 3z – 2y
⇒ 2y + 2y = x + 3z
⇒ 4y = x + 3z
⇒ x = 4y – 3z …(i)
and it is also given that x, y, z are in G.P
∴ Common ratio r = y/x = z/y …(ii)
∴ y × y = x × z
⇒ y2 = xz …(iii)
Putting the value of x = 4y – 3z in eq. (iii), we get
y2 = (4y – 3z)(z)
⇒ y2 = 4yz – 3z2
⇒ 3z2 – 4yz + y2 = 0
⇒ 3z2 – 3yz – yz + y2 = 0
⇒ 3z(z – y) – y(z – y) = 0
⇒ (3z – y)(z – y) = 0
⇒ 3z – y = 0 & z – y = 0
⇒ 3z = y & z = y but z and y are distinct numbers
z = 1/3 y & z ≠ y
z/y = 1/3
r = 1/3 [from eq. (ii)]
Hence, the correct option is (b)