Question

Let $A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-....+(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$ 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. $5$
• B. $7$
• C. $11$
• D. $9$

Answer 1

Answer: (B)

$7$

Formula: Let $a,ar,a{r}^2+a{r}^3+...+a{r}^{n-1}$ be $n$ terms of a a GP. Then its sum is given by, $S=\frac{a(1-r^n)}{1-r}$

Given, $A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-....+(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$

It is a Geometric Progression (GP) with $a=\frac{3}{4},r=\frac{-3}{4}$ and number of terms $=n$

Therefore, $A_n=\frac{\frac{3}{4}\times (1-(\frac{-3}{4}{)}^n)}{1-\left(\frac{-3}{4}\right)}$

$\Rightarrow A_n=\frac{\frac{3}{4}\times (1-(\frac{-3}{4}{)}^n)}{\frac{7}{4}}$

$\Rightarrow A_n=\frac{3}{7}[1-(\frac{-3}{4}{)}^n]$

Also given, $B_n=1-A_n$

To find: The least odd natural number $p$, such that $B_n>A_n$

Now, $1-A_n>A_n$

$\Rightarrow 1>2\times A_n$

$\Rightarrow A_n<\frac{1}{2}$

Substituting the value of $A_n$ in the above equation, we get

$\frac{3}{7}\times [1-{\left(\frac{-3}{4}\right)}^n]<\frac{1}{2}$

$\Rightarrow 1-{\left(\frac{-3}{4}\right)}^n<\frac{7}{6}$

$\Rightarrow 1-\frac{7}{6}<{\left(\frac{-3}{4}\right)}^n$

$\Rightarrow \frac{-1}{6}<{\left(\frac{-3}{4}\right)}^n$

Since $n$ is odd, then ${\left(\frac{-3}{4}\right)}^n=(-1)\times {\frac{3}{4}}^n$

Therefore, $\frac{-1}{6}<(-1)\times {\left(\frac{3}{4}\right)}^n$

Multiplying the entire inequality by $-1$, we get

$\frac{1}{6}>{\left(\frac{3}{4}\right)}^n$

Now, Applying log to the base $\frac{3}{4}$

${\log }_{\frac{3}{4}}\frac{1}{6}<\frac{3}{4}$

$\Rightarrow 6.228<n$

Therefore, $n$ should be $7$.