# Let the numbers $2,b,c$ be in an A.P. and $A=\begin{vmatrix} 1 &1 &1 \\ 2 &b &c \\ 4 &b^{2} &c^{2} \end{vmatrix}.$. If $\det (A)\epsilon \left [ 2,16 \right ],$ then c lies in the interval :  Option 1) $[2,3)$ Option 2) $\left ( 2+2^{\frac{3}{4}},4 \right )$ Option 3) $\left [ 4,6 \right ]$   Option 4) $\left [ 3,2+2^{\frac{3}{4}} \right ]$

$A=\begin{vmatrix} 1 &1 &1 \\ 2 &b &c \\ 4 & b^{2} &c^{2} \end{vmatrix}$

$C_2\rightarrow C_2-C_1;C_3-C_1$

$A=\begin{vmatrix} 1 &0 &0 \\ 2 &b-2 &c-2 \\ 4 &b^{2}-4 & c^{2}-4 \end{vmatrix}=(b-2)(c^{2}-4)-(c-2)(b^{2}-4)$

$\left | A \right |=(b-2)(c-2)(c-b)$

$2,b,c$ are in A.P

$2,2+d,2+2d\rightarrow AP.$

$\left | A \right |=(d)(2d)(d)=2d^{3}\epsilon \left [ 2,16 \right ]$

$d^{3}\equiv \left [ 1,8 \right ]$

$d\equiv \left [ 1,2 \right ]$

$C=2+2d$

$C\equiv \left [ 4,6 \right ]$

Option 1)

$[2,3)$

Option 2)

$\left ( 2+2^{\frac{3}{4}},4 \right )$

Option 3)

$\left [ 4,6 \right ]$

Option 4)

$\left [ 3,2+2^{\frac{3}{4}} \right ]$

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