1. Use the A.P. condition

Since $2,b,c$ are in A.P., the middle term is the average of the other two: $b=\frac{2+c}{2}.$

So the matrix is $A=\begin{bmatrix}1&1&1\\2&b&c\\4&b^2&c^2\end{bmatrix}.$

2. Recognize the determinant as a Vandermonde determinant

The columns are of the form $\begin{bmatrix}1\\x\\x^2\end{bmatrix}$ with $x=2,b,c$ respectively. Hence, $\det(A)=\begin{vmatrix}1&1&1\\2&b&c\\4&b^2&c^2\end{vmatrix}=(b-2)(c-2)(c-b).$

Now substitute $b=\frac{2+c}{2}.$ Then $b-2=\frac{2+c}{2}-2=\frac{c-2}{2},$ $c-b=c-\frac{2+c}{2}=\frac{c-2}{2}.$

Therefore, $\det(A)=\left(\frac{c-2}{2}\right)(c-2)\left(\frac{c-2}{2}\right)=\frac{(c-2)^3}{4}.$

3. Apply the given condition

We are given $\det(A)\in[2,16].$ So, $2\le \frac{(c-2)^3}{4}\le 16.$ Multiply throughout by $4$: $8\le (c-2)^3\le 64.$

Now take cube roots: $2\le c-2\le 4.$ Hence, $4\le c\le 6.$

So, $c\in[4,6].$

4. Check options

• A: $[2,3)$ → incorrect
• B: $[4,6]$ → correct
• C: $(2+\tfrac{23}{4},4)$ is not relevant and also malformed as an interval here
• D: $[3,2+\tfrac{23}{4}]$ → incorrect

Thus the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.