Animated Solution for Mathematics - Sequence and Series: Let, $A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-\cdots +(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$ and $B_n=1-A_n$. Then, the least odd natural number p, so that $B_n>A_n$ for all $n\ge p$, is :-

A Journey into Inequalities

Welcome, future engineer. Today, we are not just solving a problem; we are dissecting the behavior of a sequence that refuses to sit still.

When you first look at the expression $A_n=\frac{3}{4}-(\frac{3}{4}{)}^2+(\frac{3}{4}{)}^3-\cdots +(-1{)}^{n-1}(\frac{3}{4}{)}^n$, you might feel a slight hesitation. It looks like a standard sum, but those alternating signs—the plus, the minus, the plus—they tell a story of oscillation.

This is the signature of a Geometric Progression (G.P.) with a negative common ratio. Let us unravel this mystery together.

Decoding the Structure

Every complex problem is just a collection of simple truths waiting to be organized. Here, our first term is $a=\frac{3}{4}$.

The common ratio, which dictates how each term transforms into the next, is $r=-\frac{3}{4}$. Why negative? Because the signs flip with every power of $n$. If $n$ is odd, the term is positive; if $n$ is even, the term is negative.

We invoke the classic sum formula for a G.P. with $n$ terms:

Substituting our values, we get:

Look at that denominator: $1-(-\frac{3}{4})$ becomes $1+\frac{3}{4}$, which is $\frac{7}{4}$. When we divide by $\frac{7}{4}$, we are effectively multiplying by $\frac{4}{7}$.

The $4$ in the numerator and the $4$ in the denominator cancel out, leaving us with the elegant expression: This is our anchor.

The Inequality Bridge

The problem introduces $B_n=1-A_n$ and asks us to find when $B_n>A_n$. This is where many students panic, but let us stay calm.

Substitute $B_n$ with $1-A_n$. The inequality becomes $1-A_n>A_n$. Adding $A_n$ to both sides, we get $1>2A_n$, or simply $A_n<\frac{1}{2}$.

Suddenly, the problem is no longer about comparing two complex sequences; it is about finding when $A_n$ dips below the threshold of $0.5$.

Substituting our expression for $A_n$, we have:

Multiplying by $\frac{7}{3}$, we get $1-(-\frac{3}{4}{)}^n<\frac{7}{6}$. Rearranging this, we find:

This simplifies to $-(-\frac{3}{4}{)}^n<\frac{1}{6}$. Multiplying by $-1$ flips the inequality sign:

The Odd-Even Trap

Here is the crux of the problem. We are looking for the least odd natural number $p$.

When $n$ is odd, the term $(-\frac{3}{4}{)}^n$ becomes $-(\frac{3}{4}{)}^n$. Our inequality transforms into $-(\frac{3}{4}{)}^n>-\frac{1}{6}$.

Again, multiplying by $-1$ flips the sign:

Now, we are on a hunt. We need the smallest odd integer $n$ such that $(\frac{3}{4}{)}^n<0.166$. Let us test the odd numbers:

For $n=1$: $(\frac{3}{4}{)}^1=0.75$. Too large. For $n=3$: $(\frac{3}{4}{)}^3=\frac{27}{64}\approx 0.421$. Still too large. For $n=5$: $(\frac{3}{4}{)}^5=\frac{243}{1024}\approx 0.237$. Getting closer. For $n=7$: $(\frac{3}{4}{)}^7=\frac{2187}{16384}\approx 0.133$. Success!

At $n=7$, the value $0.133$ is finally less than $0.166$. We have found our target. The least odd natural number $p$ is $7$.

Mathematics is not about memorizing formulas; it is about understanding the behavior of numbers. We saw how the alternating signs created an oscillation, how the inequality simplified into a manageable threshold, and how the odd-power constraint guided our search.

You have successfully navigated the trap. Keep this analytical mindset, and no JEE problem will ever be too daunting. You are ready.