Mathematics - Sequence & Series Question with Solution | TestHub

$$A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-\dots +(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$$ Which is a G.P. with $a=\frac{3}{4},r=\frac{-3}{4}$ and number of terms $=n$ $$\begin{array}{rl}\therefore A_n& =\frac{\frac{3}{4}\times \left(1-{\left(\frac{-3}{4}\right)}^n\right)}{1-\left(\frac{-3}{4}\right)}=\frac{\frac{3}{4}\times \left(1-{\left(\frac{-3}{4}\right)}^n\right)}{\frac{7}{4}}\\ \Rightarrow A_n=\frac{3}{7}\left[1-{\left(\frac{-3}{4}\right)}^n\right]& \\ \text{ As, }{B}_n=1-A_n& \end{array}$$ 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 $$\begin{array}{rl}& \frac{3}{7}\times \left[1-{\left(\frac{-3}{4}\right)}^n\right]<\frac{1}{2}\Rightarrow 1-{\left(\frac{-3}{4}\right)}^n<\frac{7}{6}\\ & \Rightarrow 1-\frac{7}{6}<{\left(\frac{-3}{4}\right)}^n\Rightarrow \frac{-1}{6}<{\left(\frac{-3}{4}\right)}^n\end{array}$$ 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 .