Question

Let ${{A}_{n}}=\left( \dfrac{3}{4} \right)-{{\left( \dfrac{3}{4} \right)}^{2}}+{{\left( \dfrac{3}{4} \right)}^{3}}-.........+{{\left( -1 \right)}^{n-1}}{{\left( \dfrac{3}{4} \right)}^{n}}$ 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. 5
b. 7
c. 11
d. 9

Answer 1

Here, in this question, we will use the sum of the geometric progression, ${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$, we will get the condition and use the trial and error method to find the required result.
Here, we have been given ${{A}_{n}}=\left( \dfrac{3}{4} \right)-{{\left( \dfrac{3}{4} \right)}^{2}}+{{\left( \dfrac{3}{4} \right)}^{3}}-.........+{{\left( -1 \right)}^{n-1}}{{\left( \dfrac{3}{4} \right)}^{n}}$.
We can write this by taking $\left( \dfrac{3}{4} \right)$ common from every terms, so we will get, ${{A}_{n}}=\left( \dfrac{3}{4} \right)+\dfrac{3}{4}\left( \dfrac{-3}{4} \right)+\dfrac{3}{4}{{\left( \dfrac{-3}{4} \right)}^{2}}+.........+\left( \dfrac{3}{4} \right){{\left( \dfrac{3}{4} \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}$

Complete step-by-step solution:
Now, we know that when we multiply expressions with the same component, but different bases, we multiply the bases and use the same exponent.
Therefore, we get,
${{A}_{n}}=\left( \dfrac{3}{4} \right)+\dfrac{3}{4}\left( \dfrac{-3}{4} \right)+\dfrac{3}{4}{{\left( \dfrac{-3}{4} \right)}^{2}}+.........+\left( \dfrac{3}{4} \right){{\left( \dfrac{-3}{4} \right)}^{n-1}}.........(i)$
We can see that the terms are in GP, that is, geometric progression.
A GP can be as follows, $GP=a+ar+a{{r}^{2}}+a{{r}^{3}}+.........+a{{r}^{n-1}}+a{{r}^{n}}$.
Where a is the first term and r is the common ratio.
From equation (i), we get,
$a=\dfrac{3}{4},r=\left( \dfrac{-3}{4} \right),{{A}_{n}}=\text{ sum of the GP}$
We know,
Sum of GP, ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$
Now, let us substitute that values of a and r in the formula of sum to find the value of sum of GP.
Therefore, we get,
$\begin{aligned} & {{A}_{n}}=\dfrac{\dfrac{3}{4}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right)}{1-\left( \dfrac{-3}{4} \right)} \\\ & =\dfrac{\dfrac{3}{4}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right)}{1+\dfrac{3}{4}} \\\ & =\dfrac{\dfrac{3}{4}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right)}{\dfrac{4+3}{4}} \\\ & =\dfrac{\dfrac{3}{4}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right)}{\dfrac{7}{4}} \\\ & =\dfrac{3}{4}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right)\times \dfrac{4}{7} \\\ & =\dfrac{3}{7}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right).........(ii) \\\ \end{aligned}$
Now, we have,
${{B}_{n}}=1-{{A}_{n}}\text{ and }{{B}_{n}}>{{A}_{n}}$
Therefore, we can say,
$\begin{aligned} & 1-{{A}_{n}}>{{A}_{n}} \\\ & 1>{{A}_{n}}+{{A}_{n}} \\\ & 1>2{{A}_{n}} \\\ & {{A}_{n}}<\dfrac{1}{2} \\\ \end{aligned}$
From equation (ii), we get,
$\begin{aligned} & \dfrac{3}{7}\left( 1-{{\left( \dfrac{-3}{4} \right)}^{n}} \right)<\dfrac{1}{2} \\\ & 1-{{\left( \dfrac{-3}{4} \right)}^{n}}<\dfrac{7}{6} \\\ & -{{\left( \dfrac{-3}{4} \right)}^{n}}<\dfrac{7}{6}-1 \\\ & -{{\left( \dfrac{-3}{4} \right)}^{n}}<\dfrac{1}{6} \\\ \end{aligned}$
Now, let us multiply by (-) throughout the inequality, so we get,
${{\left( \dfrac{-3}{4} \right)}^{n}}>-\dfrac{1}{6}$
Let us use the trial and error method to determine the least odd number p.
First, let us find for both the cases that is, for even number and odd number.
Case 1: For even number.
We have, ${{\left( \dfrac{-3}{4} \right)}^{n}}>-\dfrac{1}{6}$
Here, we can use any even natural number for satisfying the above condition.
Case 2: For odd numbers.
We have, ${{\left( \dfrac{-3}{4} \right)}^{n}}>-\dfrac{1}{6}$
When n is an odd number, we get,

& {{\left( -1 \right)}^{n}}{{\left( \dfrac{3}{4} \right)}^{n}}>-\dfrac{1}{6} \\\ & -{{\left( \dfrac{3}{4} \right)}^{n}}>-\dfrac{1}{6} \\\ \end{aligned}$$ Now, multiply by (-) throughout the inequality. $$\begin{aligned} & {{\left( \dfrac{3}{4} \right)}^{n}}<\dfrac{1}{6} \\\ & {{\left( 0.75 \right)}^{n}}<0.167 \\\ \end{aligned}$$ We need to use the trial and error method by using n as positive odd natural numbers. When n = 1, $$\left( 0.75 \right)>0.167$$. Hence, it doesn’t satisfy the condition. When n = 2, $${{\left( 0.75 \right)}^{3}}\approx 0.422>0.167$$. Hence, it doesn’t satisfy the condition. When n = 5, $${{\left( 0.75 \right)}^{5}}\approx 0.237>0.167$$. Hence, it doesn’t satisfy the condition. When n = 7, $$\left( 0.75 \right)\approx 0.1335<0.167$$. And this satisfies the condition. Now, if we take any other values which is an odd natural number and is greater than 7, it will be satisfying the condition. Therefore, $n\ge 7$. Therefore, the least odd natural number, p that satisfies the condition for all $n\ge p,p=7$. **Hence, option (b) is the correct answer.** **Note:** When you multiply an inequality by a negative sign, the inequality changes from (>) to (<) or (<) to (>). If every term of GP is multiplied or divided by a non-zero number, then the resulting terms are also in GP.