A) \[{{P}^{3}}\]
B) \[{{P}^{2}}\]
C) \[{{P}^{10}}\]
D) \[{{P}^{4}}\]
E) \[{{P}^{5}}\]
Correct Answer: E
Solution :
Let first term and common difference of GP be a and r respectively. \[\therefore \] \[a{{r}^{2}}=P\] Now, \[a.ar.a{{r}^{2}}.a{{r}^{3}}.a{{r}^{4}}\] \[={{a}^{2}}{{r}^{10}}={{(a{{r}^{2}})}^{5}}={{p}^{5}}\]You need to login to perform this action.
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