A) \[A.P.\]
B) \[GP.\]
C) \[H.P.\]
D) None
Correct Answer: B
Solution :
\[a,\,\,b,\,\,c\]in\[A.P.\]\[\Rightarrow a+c=2b;\] \[b,\,\,c,\,\,d\]in\[G.P.\] \[\Rightarrow \] \[bd={{c}^{2}};\] \[c,\,\,d,\,\,e\]in\[H.P.\]\[\Rightarrow d=\frac{ce}{c+e}\] \[\therefore \]\[\frac{a+c}{2}\times \frac{2ce}{c+e}={{c}^{2}}\Rightarrow (a+c)e=(c+e)c\Rightarrow {{c}^{2}}=ae\]Therefore, in\[G.P.\]You need to login to perform this action.
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