Let A_n = left( frac{3}{4} right) - left( fra | vedclass

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(B) $A_n$ is a geometric progression with first term $a = \frac{3}{4}$ and common ratio $r = -\frac{3}{4}$.The sum of $n$ terms is $A_n = \frac{a(1-r^

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$7$

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(B) $A_n$ is a geometric progression with first term $a = \frac{3}{4}$ and common ratio $r = -\frac{3}{4}$.The sum of $n$ terms is $A_n = \frac{a(1-r^n)}{1-r} = \frac{\frac{3}{4}(1 - (-\frac{3}{4})^n)}{1 - (-\frac{3}{4})} = \frac{\frac{3}{4}(1 - (-\frac{3}{4})^n)}{\frac{7}{4}} = \frac{3}{7} \left[ 1 - \left( -\frac{3}{4} \right)^n \right]$.Given $B_n = 1 - A_n$,the condition $B_n > A_n$ implies $1 - A_n > A_n$,which simplifies to $1 > 2A_n$,or $A_n Substituting $A_n$: $\frac{3}{7} \left[ 1 - \left( -\frac{3}{4} \right)^n \right] This gives $-\left( -\frac{3}{4} \right)^n  -\frac{1}{6}$.For odd $n$,$\left( -\frac{3}{4} \right)^n = -\left( \frac{3}{4} \right)^n$. So,$-\left( \frac{3}{4} \right)^n > -\frac{1}{6}$,which means $\left( \frac{3}{4} \right)^n Taking logarithms: $n \log(\frac{3}{4}) $n(0.4771 - 0.6020)  \frac{0.7781}{0.1249} \approx 6.23$.The least odd natural number $p$ satisfying $n > 6.23$ is $p = 7$.

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$ such that $B_n > A_n$ for all $n \geq p$ is:

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