A) A.P.
B) G.P. only when \[x>\text{0}\]
C) G.P. if \[x<0\]
D) G.P. for all \[x\ne 0\]
Correct Answer: D
Solution :
\[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{3}}}{{{T}_{2}}}\] \[\Rightarrow {{2}^{(b-a)x}}={{2}^{(c-b)x}}\]\[\Rightarrow (b-a)x=(c-b)x\] Þ \[(b-a)=(c-b)\] \[\forall \,x,x\ne 0\] \[\therefore {{2}^{ax+1}},{{2}^{bx+1}},{{2}^{cx+1}}\]is a G.P., \[\forall \,\,x\ne 0.\]
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