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Solution

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Solution

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Solution

Determinant $\ne 0$ $\left|\begin{matrix}2 & a & 6 \ 1 & 2 & b \ 1 & 1 & 3\end{matrix}\right| \ne 0$ $=2(6-b)-a(3-b)+6(1-2)$ $=12-2b-3a+ab-6$ $=ab-3a-2b+6$ $=(a-2)(3-b)\ne 0$ $\Rightarrow a\ne2$ or $b\ne3$

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Solution

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Solution

$y = \sin(mx)$ $y_1 = m\cos(mx), \quad y_2 = -m^2\sin(mx)$ $y_3 = -m^3\cos(mx), \quad y_4 = m^4\sin(mx), \quad y_5 = m^5\cos(mx)$ $y_6 = -m^6\sin(mx), \quad y_7 = -m^7\cos(mx), \quad y_8 = m^8\sin(mx)$ $\Delta = \begin{vmatrix} \sin(mx) & m\cos(mx) & -m^2\sin(mx) \\ -m^3\cos(mx) & m^4\sin(mx) & m^5\cos(mx) \\ -m^6\sin(mx) & -m^7\cos(mx) & m^8\sin(mx) \end{vmatrix}$ $= -m^6 \begin{vmatrix} \sin(mx) & m\cos(mx) & -m^2\sin(mx) \\ -m^3\cos(mx) & m^4\sin(mx) & m^5\cos(mx) \\ \sin(mx) & m\cos(mx) & -m^2\sin(mx) \end{vmatrix}$ Since row 1 and row 3 are identical, $\Delta = 0$

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Solution

Expand determinant: $= (2-y)\left[(5-y)(10-y) - 24\right] - 2\left[2(10-y) - 18\right] + 3\left[8 - 3(5-y)\right]$ Simplify: $= (2-y)(y^2 -15y +26) + 4y -4 + 9y -21$ $= -y^3 +17y^2 -69y +52 + 13y -25$ $= -y^3 +17y^2 -56y +27$ $= 0$ Try integer root: $y=3$ $-27 +153 -168 +27 = 0$ ✔

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Solution

Expand determinant and simplify ⇒ condition reduces to $y^2 = xz$ This is condition of GP Final Answer: $\boxed{x,y,z \text{ are in GP}}$