If the system of linear equations x+2ay+az=0,    x+3by+bz=0,    x+4cy+cz=0

has a non­-zero solution, then a,b,c

  • Option 1)

    are in G.P.

  • Option 2)

    are in H.P.

  • Option 3)

    satisfy a+2b+3c=0

  • Option 4)

    are in A.P.

 

Answers (1)

As we learnt in 

Harmonic mean (HM) of two numbers a and b -

HM= \frac{2ab}{a+b}

-

 

 x+2ay+az=0

x+3by+bz=0

x+4cy+cz=0

\Rightarrow \begin{vmatrix} 1 & 2a & a\\ 1 & 3b & b\\ 1 & 4c & c \end{vmatrix}= 0

\Rightarrow \begin{vmatrix} 0 & 2a-3b & a-b\\ 0 & 3b-4c & b-c\\ 1 & 4c & c \end{vmatrix} = 0

\Rightarrow \left ( b-c \right )\left ( 2a-3b \right )-\left ( a-b \right )\left ( 3b-4c \right )= 0

\Rightarrow \left ( b-c \right )\left ( 2a-3b \right )= \left ( a-b \right )\left ( 3b-4c \right )

\Rightarrow 2ac= ab+bc

Dividing by abc we get,

\frac{2}{b}= \frac{1}{a}+ \frac{1}{c}

Therefore, a, b, c are in HP.


Option 1)

are in G.P.

This option is incorrect.

Option 2)

are in H.P.

This option is correct.

Option 3)

satisfy a+2b+3c=0

This option is incorrect.

Option 4)

are in A.P.

This option is incorrect.

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