If and an- 3-4- 3-4 2- 3-4 3- -1 n-1- 3-4 n and bn- 1-an then the smallest natural number such that bn- an- n- n0- is

The correct option is D 7 #160; B _ n =1 A _ n gt; A _ n ⇒ A _ n lt; 1 / 2 #160; Now, #160; A _ n = 3 / 4 ( 1 ( 3 / 4 #160;) # ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

If and an= ...

Question

If and $a_{n} = \frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . + {(- 1)}^{n - 1} \cdot {(\frac{3}{4})}^{n}$ and $b_{n} = 1 - a_{n}$ then the smallest natural number such that $b_{n} > a_{n} \forall n > n_{0},$ is

5

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Solution

The correct option is D $7$
$B_{n} = 1 - A_{n} > A_{n} \Rightarrow A_{n} < \frac{1}{2}$
Now, $A_{n} = \frac{\frac{3}{4} (1 - {(- \frac{3}{4})}^{n})}{1 + \frac{3}{4}} < \frac{1}{2} \Rightarrow {(- \frac{3}{4})}^{n} > - \frac{1}{6}$
Obviously, it is true for all even values of $n .$
But for
$n = 1, - \frac{3}{4} < - \frac{1}{6}$
$n = 3, {(\frac{- 3}{4})}^{3} = - \frac{27}{64} < - \frac{1}{6}$
$n = 5, {(- \frac{3}{4})}^{5} = - \frac{243}{1024} < - \frac{1}{6}$
and for $n = 7$
${(- \frac{3}{4})}^{7} = - \frac{2187}{12288} > - \frac{1}{6}$
Hence, minimum natural number $n_{0} = 7$

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Similar questions

Q. If

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 minimum natural number

n_{0}

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 $B_{n} > A_{n}, \forall n \geq n_{0}$ .

Q. 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, the least odd natural number

p

, so that

B_{n} > A_{n}

, for all

n \geq p

, is

Q. 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 :

Q. Let

a_{1} = p

and

b_{1} = q

where p and q are positive quantities. Define

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

and

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

, for even

n > 1

and

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

, for odd

n > 1

.If

p = \frac{1}{3}

and

q = \frac{2}{3}

, then what is the smallest odd n such that

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

(CAT 2007)

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If and an=34−(34)2+(34)3−...+(−1)n−1⋅(34)n and bn=1−an then the smallest natural number such that bn>an∀n>n0, is

The correct option is D 7Bn=1−An>An⇒An<12Now, An=34(1−(−34)n)1+34<12⇒(−34)n>−16Obviously, it is true for all even values of n.But for n=1,−34<−16n=3,(−34)3=−2764<−16n=5,(−34)5=−2431024<−16and for n=7(−34)7=−218712288>−16Hence, minimum natural number n0=7

If and $a_{n} = \frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . + {(- 1)}^{n - 1} \cdot {(\frac{3}{4})}^{n}$ and $b_{n} = 1 - a_{n}$ then the smallest natural number such that $b_{n} > a_{n} \forall n > n_{0},$ is