Question

If x, y, and z are in G.P. and $x+3,y+3,Z+3$ are in H.P., then

• A. y =2
• B. y =3
• C. y =1
• D. y=0

Answer 1

The correct option is B

y =3

$x,y,z$ are in G.P. Hence,

$y^2=xz$

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

$y+3=\frac{2(x+3)(z+3)}{(x+3)(z+3)}$

$=\frac{2[xz+3(x+z)+9]}{\left[\right(x+z)+6]}$

$=\frac{2[y^2+3(x+z)+9]}{[x+z+6]}$

Obviously, y=3 satisfies it.