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Properties of Iota

et A n =3/4-3...

Question

Let $A_{n} = (\frac{3}{4}) - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . + (- 1)^{n} {(\frac{3}{4})}^{n}$ and $B_{n} = 1 - A_{n}$ . Then, the least odd natural number $p$ , so that $B_{n} > A_{n}$ , for all $n \geq p$ , is :

A

9

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B

11

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C

7

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D

5

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Solution

The correct option is C $7$
$A_{n} = (\frac{3}{4}) - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . + (- 1)^{n} {(\frac{3}{4})}^{n}$
$A_{n} = \frac{\frac{3}{4} [1 - {(- \frac{3}{4})}^{n}]}{1 - (- \frac{3}{4})} = \frac{3}{7} [1 - {(- \frac{3}{4})}^{n}]$
Now, $B_{n} = 1 - A_{n}$
and $B_{n} > A_{n}$
$\Rightarrow 1 - A_{n} > A_{n}$
$\Rightarrow A_{n} < \frac{1}{2}$
$\Rightarrow \frac{3}{7} [1 - {(- \frac{3}{4})}^{n}] < \frac{1}{2}$
$\Rightarrow [1 - {(- \frac{3}{4})}^{n}] < \frac{7}{6}$
$\Rightarrow (- 1)^{n} {(\frac{3}{4})}^{n} > - \frac{1}{6}$
Let $n = p$ which is odd natural number
$\Rightarrow (- 1) {(\frac{3}{4})}^{p} > - \frac{1}{6}$
$\Rightarrow {(\frac{3}{4})}^{p} < \frac{1}{6}$
$\Rightarrow {(\frac{4}{3})}^{p} > 6$
$\Rightarrow p > \frac{ln 3 + ln 2}{2 ln 2 - ln 3}$
$\Rightarrow p > 6.23$
Hence, $n \geq p \Rightarrow n \geq 7$

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Similar questions

A_{n} = (\frac{3}{4}) - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . . + (- 1)^{n - 1} {(\frac{3}{4})}^{n}

B_{n} = 1 - A_{n}

. Then, the least odd natural number

p

B_{n} > A_{n}

n \geq p

a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + . . . + (- 1)^{n - 1} (\frac{3}{4})^{n}

b_{n} = 1 - a_{n}

then find the minimum natural number

n_{0}

b_{n} > a_{n} \forall n > n_{0}

a_{n} = \frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . + {(- 1)}^{n - 1} \cdot {(\frac{3}{4})}^{n}

b_{n} = 1 - a_{n}

then the smallest natural number such that

b_{n} > a_{n} \forall n > n_{0},

Let $A_{n} = \frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . . + (- 1)^{n - 1} {(\frac{3}{4})}^{n}$
and $B_{n} = 1 - A_{n},$ then find the least value of $n_{0}, n_{0} \in N$ such that $B_{n} > A_{n}, \forall n \geq n_{0}$ .

a_{1} = p

b_{1} = q

where p and q are positive quantities. Define

a_{n} = p b_{n - 1}

b_{n} = q b_{n - 1}

n > 1

a_{n} = p a_{n - 1}, b n = q a_{n - 1}

n > 1

p = \frac{1}{3}

q = \frac{2}{3}

, then what is the smallest odd n such that

a_{n} + b_{n} < 0.01 ?

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