Q71 / 155 · WBJEE 2018 · Mathematics

If the following three linear equations have a non-trivial solution, then $$ \begin{array}{l} x+4 a y+a z=0 \\ x+3 b y+b z=0 \\ x+2 c y+c z=0 \end{array} $$

  • A
    $a, b, c$ are in $A P$
  • B
    $a, b, c$ are in $G$
  • Correct answer
    C
    $a, b, c$ are in $\mathrm{HP}$
  • D
    $a+b+c=0$

Solution

For non-trivial solution, We have, $\left|\begin{array}{ccc}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0$ $\Rightarrow \quad 1(3 b c-2 b c)-(4 a c-2 a c)+1(4 a b-3 a b)=0$ $\Rightarrow b c-2 a c+a b=0$ $\Rightarrow b c+a b=2 a c$ $\Rightarrow b(a+c)=2 a c$ $\Rightarrow b=\frac{2 a c}{a+c}$ $\Rightarrow a, b, c$ are in $\mathrm{HP}$

Hence, the correct answer is option: 'C'.

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