A) No particular order
B) A.P.
C) G.P.
D) H.P.
Correct Answer: C
Solution :
\[a,\ b,\ c\] are in A.P. then \[2b=a+c\] .....(i) \[b,\ c,\ d\] are in G.P. then\[{{c}^{2}}=bd\] ....(ii) \[c,\ d,\ e\] are in H.P. then \[d=\frac{2ce}{c+e}\] ....(iii) From (ii), \[1,\ 2,\ 3,\ ..........=n\] \[\Rightarrow \] \[{{c}^{2}}=\frac{ace+{{c}^{2}}e}{c+e}\Rightarrow {{c}^{3}}+{{c}^{2}}e=ace+{{c}^{2}}e\] \[\Rightarrow \] \[{{c}^{3}}=ace\Rightarrow {{c}^{2}}=ae\] Hence \[a,\ c,\ e\] will be in G.P.
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