3 numbers x,y,z are in G.P,x,y2,z are in A.P., then minimum value of x+2y+z is equal to

Answers (1)

x, y, z are in G.P.

y^2 = xz    -(i)

x, y^2, z are in A.P.

y^2 = \frac{x+z}{2}    -(ii)

xz = \frac{x+z}{2}

2xz = x+z

Now, A = x+ 2y+ z

\\ = 2y^2 + 2y         -[From (i) and (ii)]

\\ \frac{dA}{dy} = 4y+2 = 0 \\ \Rightarrow y = -\frac{1}{2}

\\ \frac{d^2A}{dy^2} = 4 > 0

\therefore

 \\ A_{min}\ for\ y = -\frac{1}{2} \\ A_{min} = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) \\ = 2(\frac{1}{4})-1 \\ = -\frac{1}{2}

 

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