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(D) Given that $2, b, c$ are in $A.P.$,we can write $b = 2 + d$ and $c = 2 + 2d$,where $d$ is the common difference.The determinant is given by:$\det(

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$[4, 6]$

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(D) Given that $2, b, c$ are in $A.P.$,we can write $b = 2 + d$ and $c = 2 + 2d$,where $d$ is the common difference.The determinant is given by:$\det(A) = \begin{vmatrix} 1 & 1 & 1 \ 2 & b & c \ 4 & b^2 & c^2 \end{vmatrix}$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\det(A) = \begin{vmatrix} 1 & 0 & 0 \ 2 & b-2 & c-2 \ 4 & b^2-4 & c^2-4 \end{vmatrix}$Expanding along the first row:$\det(A) = (b-2)(c^2-4) - (c-2)(b^2-4)$$= (b-2)(c-2)(c+2) - (c-2)(b-2)(b+2)$$= (b-2)(c-2)(c+2 - b - 2) = (b-2)(c-2)(c-b)$Substituting $b = 2+d$ and $c = 2+2d$:$\det(A) = (d)(2d)(2d - d) = (d)(2d)(d) = 2d^3$Given $\det(A) \in [2, 16]$,we have $2 \le 2d^3 \le 16$,which implies $1 \le d^3 \le 8$,so $1 \le d \le 2$.Since $c = 2 + 2d$,we have $2(1) + 2 \le c \le 2(2) + 2$,which gives $4 \le c \le 6$.Thus,$c \in [4, 6]$.

Let the numbers $2, b, c$ be in an $A.P.$ and $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix}$. If $\det(A) \in [2, 16]$,then $c$ lies in the interval:

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