If a, b, c are in AP , then what is | array lll x+1 & x+2 & x+3 x+2 & x+3 & x+4 x+a & x+b & x+3 array | equal to?

Correct answer: B. 0 | array lll x+1 & x+2 & x+3 x+2 & x+3 & x+4 x+a & x+b & x+3 array | Here a_ 33 element is wrong it must be x+c a, b, c are in AP 2 b=a+cC_2 arrow C_2-C_1 and C_3 arrow C_3-C_2| array ccc x+1 & 1 & 1 x+2 & 1 & 1 x+a & b-a & c-b array | b-a=c-b Column C_2 and C_3 are identical. Value of determinant =0 .

NDA 2023

Exam: NDA
Year: 2023
Subject: Mathematics
Chapter: Determinants

NDA 2023 · Mathematics · Determinants

If a, b, c are in AP , then what is | array lll x+1 & x+2 & x+3 x+2...

a, b, c

are in AP , then what is

\left|

\begin{array}{lll}x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+3\end{array}

\right|

equal to?$

Options

A $-1$
B $0$
C $1$
D $2$

Correct answer

B. $0$

Step-by-step solution

\left|

\begin{array}{lll}x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+3\end{array}

\right|

Here

a_{33}

element is wrong it must be

x+c\because \quad a, b, c

are in AP

2 b=a+cC_2 \rightarrow C_2-C_1

and

C_3 \rightarrow C_3-C_2\left|

\begin{array}{ccc}x+1 & 1 & 1 \\ x+2 & 1 & 1 \\ x+a & b-a & c-b\end{array}

\right|\because \quad b-a=c-b\therefore \quad

Column

C_2

and

C_3

are identical.

\therefore \quad

Value of determinant

=0

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