KCET · Maths · Determinants
If \(\left|\begin{array}{lll}\mathrm{x}+1 & \mathrm{x}+2 & \mathrm{x}+\mathrm{a} \\ \mathrm{x}+2 & \mathrm{x}+3 & \mathrm{x}+\mathrm{b} \\ \mathrm{x}+3 & \mathrm{x}+4 & \mathrm{x}+\mathrm{c}\end{array}\right|=0\), then \(a, b, c\) are
- A in GP
- B in HP
- C equal
- D in AP
Answer & Solution
Correct Answer
(D) in AP
Step-by-step Solution
Detailed explanation
Given, \(\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0\)
Applying \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\), we get
\[
\begin{aligned}
&\quad\left|\begin{array}{ccc}
0 & 0 & a+c-2 b \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|=0 \\
&\Rightarrow(a+c-2 b)\left[x^{2}+6 x+8-\left(x^{2}+6 x+9\right)\right] \\
&\Rightarrow \quad(a+c-2 b)(-1)=0 \\
&\Rightarrow \quad 2 b=a+c \\
&\Rightarrow \quad a, b, c \text { are in AP. }
\end{aligned}
\]
Applying \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\), we get
\[
\begin{aligned}
&\quad\left|\begin{array}{ccc}
0 & 0 & a+c-2 b \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|=0 \\
&\Rightarrow(a+c-2 b)\left[x^{2}+6 x+8-\left(x^{2}+6 x+9\right)\right] \\
&\Rightarrow \quad(a+c-2 b)(-1)=0 \\
&\Rightarrow \quad 2 b=a+c \\
&\Rightarrow \quad a, b, c \text { are in AP. }
\end{aligned}
\]
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