A) A.P.
B) \[G.P.\text{ }only\text{ }when\,\,\,x>0\]
C) \[\operatorname{G}.P. if;\,\,x<0\]
D) \[\operatorname{G}.P.for\,\,x\ne 0\]
Correct Answer: D
Solution :
Let, \[{{T}_{1}}={{2}^{ax+1}}\] \[{{T}_{2}}={{2}^{bx+1}}\] and \[{{T}_{3}}=\,\,{{2}^{cx+1}}\] \[\because \,\,\,\frac{{{T}_{2}}}{{{T}_{1}}}={{2}^{(b-a)x}}\,\,and\,\,\frac{{{T}_{3}}}{{{T}_{2}}}={{2}^{(c-b)x}}\] \[\therefore \,\,\,\,\,\,\frac{{{\Tau }_{2}}}{{{T}_{1}}}=\frac{{{T}_{3}}}{{{T}_{2}}}\] \[\because \,\,\,\,(b\,-a)=\,\,(c\,-b)\forall \,\,x,\,\,x\ne 0\] Hence, \[{{2}^{ax+1}},\,\,{{2}^{bx+1}},\,\,{{2}^{cx+1}}\,are\,\,in\,\,G.P.,\,\,\forall \,x\ne 0\]
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