A) are in AP
B) are in GP
C) are in HP
D) satisfy a+2b+3c=0
Correct Answer: C
Solution :
The system of linear equations \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z=0\] \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z=0\] and \[{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z=0\] has a non-zero solution , if \[\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0\]. The system of linear equations has a non-zero solution, then \[\left| \begin{matrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \\ \end{matrix} \right|=0\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\left| \begin{matrix} 1 & 2a & a \\ 0 & 3b-2a & b-a \\ 0 & 4c-2a & c-a \\ \end{matrix} \right|=0\] \[\Rightarrow \,\,\,(3b-2a)\,(c-a)-(4c-2a)(b-a)=0\] \[\Rightarrow 3bc-3ba-2ac+2{{a}^{2}}=4bc-2ab\] \[-4ac+2{{a}^{2}}\] \[\Rightarrow 4ac-2ac=4bc-2ab-3bc+3ab\] \[\Rightarrow \] \[2ac=bc+ab\] On dividing by abc, we get \[\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\] i.e., a, b, c are in HP.
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