Which is a $G.P.$ with $a = \frac{3}{4}'r = \frac{{ - 3}}{4}$ and number of terms $=n$
${A_n} = \frac{{\frac{3}{4} \times \left( {1 - {{\left( {\frac{{ - 3}}{4}} \right)}^n}} \right)}}{{1 - \left( {\frac{{ - 3}}{4}} \right)}} - \frac{{\frac{3}{4} \times \left( {1 - {{\left( {\frac{{ - 3}}{4}} \right)}^n}} \right)}}{{\frac{7}{4}}}$
$ \Rightarrow {A_n} = \frac{3}{7}\left[ {1 - {{\left( {\frac{{ - 3}}{4}} \right)}^n}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)$
As, ${B_n} = 1 - {A_n}$
For least odd natural number $p$, such that ${B_n} > {A_n}$
$ \Rightarrow 1 - {A_n} > {A_n}\,\,\,\,\, \Rightarrow 1 > 2 > \times {A_n}\,\,\,\,\, \Rightarrow {A_n} < \frac{1}{2}$
From eqn. $(1)$, we get
$\frac{3}{7} \times \left[ {1 - {{\left( {\frac{{ - 3}}{4}} \right)}^n}} \right] < \frac{1}{2}\,\, \Rightarrow 1 - {\left( {\frac{{ - 3}}{4}} \right)^n} < \frac{7}{6}$
$ \Rightarrow 1 - \frac{7}{6} < {\left( {\frac{{ - 3}}{4}} \right)^n} \Rightarrow \frac{{ - 1}}{6} < {\left( {\frac{{ - 3}}{4}} \right)^n}$
As $n$ is odd, then ${\left( {\frac{{ - 3}}{4}} \right)^n} = - \frac{{{3^n}}}{4}$
So $\frac{{ - 1}}{6} < - {\left( {\frac{3}{4}} \right)^n}\,\,\, \Rightarrow \frac{1}{6} > {\left( {\frac{3}{4}} \right)^n}$
$\log \left( {\frac{1}{6}} \right) = n\,\log \left( {\frac{3}{4}} \right) \Rightarrow 6.228 < n$
Hence, $n$ should be $7$.
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