A) HP
B) GP
C) AP
D) none of these
Correct Answer: A
Solution :
Since, a, b, c are in AP, then \[2b=a+\text{ }c\] ...(i) and \[a,\text{ }mb,\text{ }c\]are in GP, then \[mb=\sqrt{ac}\] or \[{{m}^{2}}{{b}^{2}}=ac\] ?(ii) From Eqs. (i) and (ii) \[\frac{{{m}^{2}}{{b}^{2}}}{2b}=\frac{ac}{a+c}\] \[\Rightarrow \] \[{{m}^{2}}b=\frac{2ac}{a+c}\] \[\Rightarrow \] \[a,{{m}^{2}}b,c\]are in HP.You need to login to perform this action.
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