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# Question and Answer

If a, b, c, are in A.P, then the determinant
$$\left|\begin{array}{ccc} {x+2} & {x+3} & {x+2 a} \\ {x+3} & {x+4} & {x+2 b} \\ {x+4} & {x+5} & {x+2 c} \end{array}\right|$$ is
• 1
• 2x
• x
• 0
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## QuestionMathsClass 12

If a, b, c, are in A.P, then the determinant
$$\left|\begin{array}{ccc} {x+2} & {x+3} & {x+2 a} \\ {x+3} & {x+4} & {x+2 b} \\ {x+4} & {x+5} & {x+2 c} \end{array}\right|$$ is
• 1
• 2x
• x
• 0

Let $$\Delta=\left|\begin{array}{lll} {x+2} & {x+3} & {x+2 a} \\ {x+3} & {x+4} & {x+2 b} \\ {x+4} & {x+5} & {x+2 c} \end{array}\right|$$
Since a, b, c are in A.P.
$$\therefore$$ 2b = a + c
$$\Delta=\left|\begin{array}{ccc} {x+2} & {x+3} & {x+2 a} \\ {x+3} & {x+4} & {x+(a+c)} \\ {x+4} & {x+5} & {x+2 c} \end{array}\right|$$
Applying Elementary Row transformations
$$R_1 \rightarrow R_1 - R_2,~~ and~~ R_3\rightarrow R_3 - R_2$$
$$\Delta=\left|\begin{array}{ccc} {-1} & {-1} & {a-c} \\ {\mathrm{x}+3} & {\mathrm{x}+4} & {\mathrm{x}+\mathrm{a}+\mathrm{c}} \\ {1} & {1} & {\mathrm{c}-\mathrm{a}} \end{array}\right|$$
$$R_1 \rightarrow R_1+R_3$$
$$\Delta=\left|\begin{array}{ccc} {0} & {0} & {0} \\ {\mathrm{x}+3} & {\mathrm{x}+4} & {\mathrm{x}+\mathrm{a}+\mathrm{c}} \\ {1} & {1} & {\mathrm{c}-\mathrm{a}} \end{array}\right|$$
[In a determinant if all elements of a row is 0 then the value of determinant is 0]
So, here all the elements of first row ($$R_1$$) are zero.
$$\therefore$$ $$\Delta=0$$
Therefore the choice is: D