Question

Let a, x, b be in A.P.; a, y, b in G.P. and a, z, b in H.P. where a and b are distinct positive real numbers. If x = y + 2 and a = 5z, then

• A. $y^2=zx$
• B. $x>y>z$
• C. $a=9,b=1$
• D. $a=\frac{1}{4},b=\frac{-9}{4}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$y^2=zx$

B

$x>y>z$

C

$a=9,b=1$

$a,x,binA.P\Rightarrow 2x=a+ba,y,binG.P\Rightarrow y^2=aba,z,binH.P\Rightarrow z=\frac{2ab}{a+b}\therefore x>y>z(A>G>H)$
At $y^2=xz$
$\therefore a=5z$ (given) ..... (i)
$\frac{a}{5}=\frac{2ab}{a+b}$ or $a=9b$
x = y + 2 .... (ii)
or $\frac{a+b}{2}=\sqrt{ab}+2$
or 5b = 3b + 2
b = 1 and a = 9