A) G.P. with common ratio \[{{e}^{d}}\]
B) G.P. with common ratio \[{{e}^{1/d}}\]
C) G.P. with common ratio \[{{e}^{d/({{b}^{2}}-{{d}^{2}})}}\]
D) A.P.
Correct Answer: C
Solution :
[c] a, b, c are in A.P. \[\Rightarrow \,\,\,\,2b=a+c\] Now, \[{{e}^{1/c}}\times {{e}^{1/a}}={{e}^{(a+c)/ac}}={{e}^{2b/ac}}={{({{e}^{b/ac}})}^{2}}\] \[\therefore \,\,\,{{e}^{1/c}},\,\,{{e}^{b/ac}},{{e}^{1/a}}\] in GP. with common ratio \[=\frac{{{e}^{b/ac}}}{{{e}^{1/c}}}={{e}^{(b-a)/ac}}={{e}^{d/(b-d)(b+d)}}\] \[={{e}^{d/({{b}^{2}}-{{d}^{2}})}}\] [ \[\because \] a, b, c are in A.P. with common difference d \[\therefore \,\,\,\,b-a=c-b=d\]]
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