A) \[0\]
B) \[1\]
C) \[-1\]
D) None of these
Correct Answer: A
Solution :
[a] Since, a, b, c are in GP. \[\Rightarrow \,{{b}^{2}}=ac\] Expanding the determinant we get, \[\left| \begin{matrix} a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0 \\ \end{matrix} \right|\] \[=a\left| \begin{matrix} c & b+c \\ b+c & 0 \\ \end{matrix} \right|-b\left| \begin{matrix} b & b+c \\ a+b & 0 \\ \end{matrix} \right|\] \[+(a+b)\left| \begin{matrix} b & c \\ a+b & b+c \\ \end{matrix} \right|\] \[=-a{{(b+c)}^{2}}+b(a+b)(b+c)+(a+b)\] \[({{b}^{2}}+bc-ac-bc)\] \[=-a({{b}^{2}}+{{c}^{2}}+2bc)+b(ab+ac+{{b}^{2}}+bc)\] \[=-a{{b}^{2}}-a{{c}^{2}}-2abc+a{{b}^{2}}+2abc+{{b}^{2}}c\] \[(\because {{b}^{2}}=ac)\] \[=-a{{c}^{2}}+{{b}^{2}}c=-a{{c}^{2}}+ac.c=-a{{c}^{2}}+a{{c}^{2}}=0\]
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