$$
A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n
$$
Which is a G.P. with $a=\frac{3}{4}, r=\frac{-3}{4}$ and number of terms $=n$
$$
\begin{aligned}
\therefore A_n &=\frac{\frac{3}{4} \times\left(1-\left(\frac{-3}{4}\right)^n\right)}{1-\left(\frac{-3}{4}\right)}=\frac{\frac{3}{4} \times\left(1-\left(\frac{-3}{4}\right)^n\right)}{\frac{7}{4}} \\
\Rightarrow A_n=\frac{3}{7}\left[1-\left(\frac{-3}{4}\right)^n\right] \\
\text { As, } B_n=1-A_n
\end{aligned}
$$
For least odd natural number $\mathrm{p}$, 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
$$
\begin{aligned}
&\frac{3}{7} \times\left[1-\left(\frac{-3}{4}\right)^n\right] < \frac{1}{2} \Rightarrow 1-\left(\frac{-3}{4}\right)^n < \frac{7}{6} \\
&\Rightarrow 1-\frac{7}{6} < \left(\frac{-3}{4}\right)^n \Rightarrow \frac{-1}{6} < \left(\frac{-3}{4}\right)^n
\end{aligned}
$$
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 .