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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $A_n$ is a geometric progression $(G.P.)$ with first term $a = \frac{3}{4}$,common ratio $r = -\frac{3}{4}$,and $n$ terms.Using the sum formula $S

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) $A_n$ is a geometric progression $(G.P.)$ with first term $a = \frac{3}{4}$,common ratio $r = -\frac{3}{4}$,and $n$ terms.Using the sum formula $S_n = \frac{a(1-r^n)}{1-r}$,we get:$A_n = \frac{\frac{3}{4} \left( 1 - (-\frac{3}{4})^n \right)}{1 - (-\frac{3}{4})} = \frac{\frac{3}{4} \left( 1 - (-\frac{3}{4})^n \right)}{\frac{7}{4}} = \frac{3}{7} \left[ 1 - (-\frac{3}{4})^n \right] \quad (1)$Given $B_n = 1 - A_n$,the condition $B_n > A_n$ implies:$1 - A_n > A_n \implies 1 > 2A_n \implies A_n Substituting $(1)$ into the inequality:$\frac{3}{7} \left[ 1 - (-\frac{3}{4})^n \right] $-(-\frac{3}{4})^n  -\frac{1}{6}$Since we are looking for an odd natural number $p$,let $n$ be odd. Then $(-\frac{3}{4})^n = -(\frac{3}{4})^n$. The inequality becomes:$-(\frac{3}{4})^n > -\frac{1}{6} \implies (\frac{3}{4})^n Taking logarithms: $n \log(\frac{3}{4}) $n (\log 3 - \log 4) $n (-0.1249)  \frac{0.7781}{0.1249} \approx 6.228$The least odd natural number $n \geq p$ satisfying this is $7$.

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

Similar Questions

If the first term of a $G.P.$ $a_1, a_2, a_3, \dots$ is unity such that $4a_2 + 5a_3$ is least, then the common ratio of the $G.P.$ is

The sum of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ up to $20$ terms is

If $a_1, a_2, a_3, ..., a_n$ are in $A.P.$,where $a_i > 0$ for all $i$,then the value of $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = $

If $f(x + y, x - y) = xy$,then the arithmetic mean of $f(x, y)$ and $f(y, x)$ is

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module