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JEE Main 2003
Exam
JEE Main
Year
2003
Subject
Mathematics
Chapter
Determinants

JEE Main 2003 · Mathematics · Determinants

If the system of linear equations x+2 a y+a z=0 x+3 b y+b z=0 x+4 c...

If the system of linear equations x+2ay+az=0x+2 a y+a z=0 x+3by+bz=0x+3 b y+b z=0 x+4cy+cz=0x+4 c y+c z=0 has a non-zero solution, then a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c}

Options

  1. Asatisfy a+2b+3c=0a+2 b+3 c=0
  2. Bare in A.P.
  3. Care in G.P.
  4. Dare in H.P.

Correct answer

D. are in H.P.

Step-by-step solution

\left|12aa13 b b14cc\begin{array}{lll}1 & 2 \mathrm{a} & \mathrm{a} \\ 1 & 3 \mathrm{~b} & \mathrm{~b} \\ 1 & 4 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{C}_2 \rightarrow \mathrm{C}_2-2 \mathrm{C}_3\left|10a1 b b12cc\begin{array}{ccc}1 & 0 & \mathrm{a} \\ 1 & \mathrm{~b} & \mathrm{~b} \\ 1 & 2 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1\left|10a0bb−a02c−bc−b\begin{array}{ccc}1 & 0 & a \\ 0 & b & b-a \\ 0 & 2 c-b & c-b\end{array}\right|=0b(c-b)-(b-a)(2 c-b)=0 On simplification, 2 b=1a+1c∴a,b,c\frac{2}{\mathrm{~b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}\therefore \mathrm{a}, \mathrm{b}, \mathrm{c} are in Harmonic Progression.
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