If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.

1.	If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P. then the common ratio of the G.P. is A. 3 B. 1/3 C. 2 D. 1/2
Answer» 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)

If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P. then the common ratio of the G.P. is A. 3 B. 1/3 C. 2 D. 1/2

Answer»

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)