Let a, b, c, d be in arithmetic progression w | vedclass

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(B) Given that $a, b, c, d$ are in arithmetic progression with common difference $\lambda$,we have $b = a + \lambda$,$c = a + 2\lambda$,and $d = a + 3

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$1$

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(B) Given that $a, b, c, d$ are in arithmetic progression with common difference $\lambda$,we have $b = a + \lambda$,$c = a + 2\lambda$,and $d = a + 3\lambda$.Substituting these into the determinant:$a - c = -2\lambda$,$b - d = -2\lambda$,$d - b = 2\lambda$,$c - b = \lambda$,$d - c = \lambda$.The determinant is $\Delta = \left|\begin{array}{lll} x-2\lambda & x+b & x+a \ x-1 & x+c & x+b \ x+2\lambda & x+d & x+c \end{array}ight| = 2$.Applying column operations $C_2 ightarrow C_2 - C_3$:$\Delta = \left|\begin{array}{lll} x-2\lambda & \lambda & x+a \ x-1 & \lambda & x+b \ x+2\lambda & \lambda & x+c \end{array}ight| = 2$.Taking $\lambda$ common from $C_2$:$\Delta = \lambda \left|\begin{array}{lll} x-2\lambda & 1 & x+a \ x-1 & 1 & x+b \ x+2\lambda & 1 & x+c \end{array}ight| = 2$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta = \lambda \left|\begin{array}{lll} x-2\lambda & 1 & x+a \ 2\lambda-1 & 0 & \lambda \ 4\lambda & 0 & 2\lambda \end{array}ight| = 2$.Expanding along $C_2$:$\Delta = \lambda \cdot (-1) \cdot ((2\lambda-1)(2\lambda) - (4\lambda)(\lambda)) = 2$.$\Delta = -\lambda \cdot (4\lambda^2 - 2\lambda - 4\lambda^2) = 2$.$\Delta = -\lambda \cdot (-2\lambda) = 2\lambda^2 = 2$.Therefore,$\lambda^2 = 1$.

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If
$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$
then the value of $\lambda^{2}$ is equal to $.....$

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Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$then the value of $\lambda^{2}$ is equal to $.....$