#### Explain Solution R. D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 47 maths Textbook Solution.

Hint $a, b, c \text { are in AP }$

Given: $\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0 \text { where } a, b, c \text { are in } \mathrm{AP}$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right| \\ &\text { Now } R_{3} \rightarrow-2 R_{2}+R_{1}+R_{3} \\ &=\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ 0 & 0 & a+c-2 b \end{array}\right| \end{aligned}

As $a b_{2} c \text { are in } \mathrm{AP}, 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$

Substituting the value in the denominator,

\begin{aligned} &=\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ 0 & 0 & a+c-a-c \end{array}\right| \\ &=\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ 0 & 0 & 0 \end{array}\right| \end{aligned}

Expanding along $R_{3}$

$=0$