JEE Main 2018 — Mathematics Algebra

$$A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n$$An​=(43​)−(43​)2+(43​)3−…+(−1)n−1(43​)n Which is a G.P. with $a=\frac{3}{4}, r=\frac{-3}{4}$a=43​,r=4−3​ and number of terms $=n$=n $$\begin{aligned} \therefore A_n &=\frac{\frac{3}{4} \times\left(1-\left(\frac{-3}{4}\right)^n\right)}{1-\left(\frac{-3}{4}\right)}=\frac{\frac{3}{4} \times\left(1-\left(\frac{-3}{4}\right)^n\right)}{\frac{7}{4}} \\ \Rightarrow A_n=\frac{3}{7}\left[1-\left(\frac{-3}{4}\right)^n\right] \\ \text { As, } B_n=1-A_n \end{aligned}$$∴An​⇒An​=73​[1−(4−3​)n] As, Bn​=1−An​​=1−(4−3​)43​×(1−(4−3​)n)​=47​43​×(1−(4−3​)n)​ For least odd natural number $\mathrm{p}$p, such that $B_n>A_n$Bn​>An​ $$\Rightarrow 1-A_n>A_n \Rightarrow 1>2 \times A_n \Rightarrow A_n < \frac{1}{2}$$⇒1−An​>An​⇒1>2×An​⇒An​<21​ From eqn. (1), we get $$\begin{aligned} &\frac{3}{7} \times\left[1-\left(\frac{-3}{4}\right)^n\right] < \frac{1}{2} \Rightarrow 1-\left(\frac{-3}{4}\right)^n < \frac{7}{6} \\ &\Rightarrow 1-\frac{7}{6} < \left(\frac{-3}{4}\right)^n \Rightarrow \frac{-1}{6} < \left(\frac{-3}{4}\right)^n \end{aligned}$$​73​×[1−(4−3​)n]<21​⇒1−(4−3​)n<67​⇒1−67​<(4−3​)n⇒6−1​<(4−3​)n​ As $n$n is odd, then $\left(\frac{-3}{4}\right)^n=-\frac{3^n}{4}$(4−3​)n=−43n​ So $\frac{-1}{6} < -\left(\frac{3}{4}\right)^n \Rightarrow \frac{1}{6}>\left(\frac{3}{4}\right)^n$6−1​<−(43​)n⇒61​>(43​)n $$\log \left(\frac{1}{6}\right)=n \log \left(\frac{3}{4}\right) \Rightarrow 6.228 < n$$log(61​)=nlog(43​)⇒6.228<n Hence, $n$n should be 7 .