A) \[a\ne b\ne c\]
B) \[{{a}^{2}}={{b}^{2}}=\frac{{{c}^{2}}}{2}\]
C) \[a,\,b,\,c\] are in G.P.
D) \[\frac{-a}{2},b,c\]are in G.P
Correct Answer: D
Solution :
a, b, c, are in A.P. Þ 2b = a + c,b - a = c - b \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in H.P. \[\frac{1}{{{b}^{2}}}-\frac{1}{{{a}^{2}}}=\frac{1}{{{c}^{2}}}-\frac{1}{{{b}^{2}}}\] \[\Rightarrow \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}=\frac{{{b}^{2}}-{{c}^{2}}}{{{b}^{2}}{{c}^{2}}}\] Þ \[(a-b)[{{c}^{2}}(a+b)-{{a}^{2}}(b+c)]=0\], \[[\because \,(b-c)=(a-b)]\] Þ \[a=b\] or \[{{c}^{2}}a+{{c}^{2}}b-{{a}^{2}}b-{{a}^{2}}c=0\] Þ \[{{c}^{2}}a+{{c}^{2}}b-{{a}^{2}}b-{{a}^{2}}c=0\] Þ \[ac\,(c-a)=b\,({{a}^{2}}-{{c}^{2}})\] Þ \[ac=-b\,(c+a)\] Þ \[-ac=b.2b\] Þ \[{{b}^{2}}=(-a/2)\,c\], \[\therefore -a/2,b,c\] are in G.P.
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