A) 5
B) 7
C) 11
D) 9
Correct Answer: B
Solution :
Formula: Let \[a,ar,a{{r}^{2}}+a{{r}^{3}}+....+a{{r}^{n-1}}\]be\[n\] terms of a a GP. Then its sum is given by, \[S=\frac{a(1-{{r}^{n}})}{1-r}\] Given, \[{{A}_{n}}=\left( \frac{3}{4} \right)-{{\left( \frac{3}{4} \right)}^{2}}+{{\left( \frac{3}{4} \right)}^{3}}-......+{{(-1)}^{n-1}}{{\left( \frac{3}{4} \right)}^{n}}\] It is a Geometric Progression (GP) with \[a=\frac{3}{4},r=\frac{-3}{4}\]and number of terms \[=n\] Therefore, \[{{A}_{n}}=\frac{\frac{3}{4}\times (1-{{(\frac{-3}{4})}^{n}})}{1-(\frac{-3}{4})}\] \[\Rightarrow {{A}_{n}}=\frac{\frac{3}{4}\times (1-{{(\frac{-3}{4})}^{n}})}{\frac{7}{4}}\] \[\Rightarrow {{A}_{n}}=\frac{3}{7}\left[ 1-{{(\frac{-3}{4})}^{n}} \right]\] Also given, \[{{B}_{n}}=1-{{A}_{n}}\] To find: The least odd natural number \[p\], such that \[{{B}_{n}}>{{A}_{n}}\] Now, \[1-{{A}_{n}}>{{A}_{n}}\] \[\Rightarrow {{A}_{n}}<\frac{1}{2}\] Substituting the value of \[{{A}_{n}}\] in the above equation, we get \[\frac{3}{7}\times \left[ 1-{{\left( \frac{-3}{4} \right)}^{n}} \right]<\frac{1}{2}\] \[\Rightarrow 1-\left( \frac{-3}{4} \right)<\frac{7}{6}\] \[\Rightarrow 1-\frac{7}{6}<\left( \frac{-3}{4} \right)\] \[\Rightarrow \frac{-1}{6}<{{\left( \frac{-3}{4} \right)}^{n}}\] Since \[n\] is odd, then\[{{\left( \frac{-3}{4} \right)}^{n}}=(-1)\times \frac{{{3}^{n}}}{4}\] Therefore, \[\frac{-1}{6}<(-1)\times {{\left( \frac{3}{4} \right)}^{n}}\] Multiplying the entire inequality by \[-1\], we get \[\frac{1}{6}>{{\left( \frac{3}{4} \right)}^{n}}\] Now, Applying log to the base\[\frac{3}{4}\] \[{{\log }_{\frac{3}{4}}}\frac{1}{6}<\frac{3}{4}\] \[\Rightarrow 6.228<n\] Therefore, \[n\] should be 7.
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