A) The (n + 1)th power of the nth term of the GP
B) The (2n + 1)th power of the nth term of the GP
C) The (2n + 1)th power of the (n + 1)th term of the GP
D) The nth power of the (n + 1)th terms of the GP
Correct Answer: C
Solution :
[c] The GP is a, \[ar,a{{r}^{2}},......a{{r}^{2n}}\] So, \[P=a.\,ar..a{{r}^{2}}.a{{r}^{3}}........a{{r}^{2n}}\] \[={{a}^{2n+1}}.{{r}^{1+2+......+2n}}\] \[={{a}^{(2n+1)}}r\,{{\,}^{\frac{2n(2n+1)}{2}}}={{a}^{2n+1}}{{r}^{n(2n+1)}}={{(a{{r}^{n}})}^{(2n+1)}}\] \[=(2n+1)th\] power of the \[(n+1)th\] term of GP.You need to login to perform this action.
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