Q100 / 180 · KCET 2009 · Mathematics

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
  • Correct answer
    D
    in AP

Solution

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 $$ \begin{aligned} &\quad\left|\begin{array}{ccc} 0 & 0 & a+c-2 b \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0 \\ &\Rightarrow(a+c-2 b)\left[x^{2}+6 x+8-\left(x^{2}+6 x+9\right)\right] \\ &\Rightarrow \quad(a+c-2 b)(-1)=0 \\ &\Rightarrow \quad 2 b=a+c \\ &\Rightarrow \quad a, b, c \text { are in AP. } \end{aligned} $$

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

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