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If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}$, then what is the value of
$\left|\begin{array}{lll}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$
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$\left|\begin{array}{lll}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$
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Since, $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}$.
$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
Explanding the determinant we get,
$\left|\begin{array}{ccc}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$
$=a\left|\begin{array}{cc}c & b+c \\ b+c & 0\end{array}\right|-b\left|\begin{array}{cc}b & b+c \\ a+b & 0\end{array}\right|+(a+b)\left|\begin{array}{cc}b & c \\ a+b & b+c\end{array}\right|$
$=-a(b+c)^{2}+b(a+b)(b+c)+(a+b)\left(b^{2}+b c-a c-b c\right)$
$=-a\left(b^{2}+c^{2}+2 b c\right)+b\left(a b+a c+b^{2}+b c\right)$
$=-a b^{2}-a c^{2}-2 a b c+a b^{2}+2 a b c+b^{2} c \quad\left(\because b^{2}=a c\right)$
$=-a c^{2}+b^{2} c=-a c^{2}+a c \cdot c .$
$=-a c^{2}+a c^{2}=0$
$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
Explanding the determinant we get,
$\left|\begin{array}{ccc}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$
$=a\left|\begin{array}{cc}c & b+c \\ b+c & 0\end{array}\right|-b\left|\begin{array}{cc}b & b+c \\ a+b & 0\end{array}\right|+(a+b)\left|\begin{array}{cc}b & c \\ a+b & b+c\end{array}\right|$
$=-a(b+c)^{2}+b(a+b)(b+c)+(a+b)\left(b^{2}+b c-a c-b c\right)$
$=-a\left(b^{2}+c^{2}+2 b c\right)+b\left(a b+a c+b^{2}+b c\right)$
$=-a b^{2}-a c^{2}-2 a b c+a b^{2}+2 a b c+b^{2} c \quad\left(\because b^{2}=a c\right)$
$=-a c^{2}+b^{2} c=-a c^{2}+a c \cdot c .$
$=-a c^{2}+a c^{2}=0$
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