Question

If x, y, z are in A.P. and x, y, (z + 1) are in G.P. then

Answer 1

Answer:

Correct option is B)

Given,

x,y,z are in G.P.

∴y

2

=xz

Now, x+3,y+3,z+3 are in H.P.

∴y+3=

(x+3)+(z+3)

2(x+3)(z+3)

⇒y+3=

[(x+z)+6]

2[xz+3(x+z)+9]

⇒y+3=

[x+z+6]

2[y

2

+3(x+z)+9

⇒(y+3)(x+z+6)=2y

2

+6(x+z)+18

⇒2y

2

−(x+z+6)y+3(x+z)=0...(i)

Substituting the values of y from given options, we get

y=3 satisfies equation (i).