The value of 12-34-56-78-99100lies in the interval-

Let X=12·34·56·78⋯99100 ⇒ X lt;23·45·67·89⋯100101 as nn+1 lt;n+1n+2 for nbsp;n gt;0 ⇒ X^2 lt;12·23·34·45⋯99100·100101=1101 lt;1100 ⇒ X lt;110 Also ...

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Standard VI

Mathematics

Idea of a Set

The value of ...

Question

The value of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \dots \frac{99}{100}$
lies in the interval,

(\frac{1}{15}, \frac{1}{10})

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(\frac{1}{20}, \frac{1}{15})

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(\frac{1}{10}, \frac{1}{5})

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(\frac{1}{10}, 1)

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Solution

The correct option is A $(\frac{1}{15}, \frac{1}{10})$
Let $X = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \dots \frac{99}{100}$
$\Rightarrow X < \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{8}{9} \dots \frac{100}{101}$ as
$\frac{n}{n + 1} < \frac{n + 1}{n + 2}$ for $n > 0$
$\Rightarrow X^{2} < \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots \frac{99}{100} \cdot \frac{100}{101} = \frac{1}{101} < \frac{1}{100}$
$\Rightarrow X < \frac{1}{10}$
Also $x >< \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots \frac{98}{99} \cdot \frac{99}{100} = \frac{1}{200} < \frac{1}{225}$
$\Rightarrow X < \frac{1}{15}$
$∴ \frac{1}{15} < X < \frac{1}{10} \Rightarrow X \in (\frac{1}{15}, \frac{1}{10})$

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Similar questions

Q. The value of

\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \dots \frac{99}{100}

lies in the interval,

The following are the marks obtained in Maths by students of Class 8:
51, 94, 34, 20, 50, 81, 81, 90, 76, 34, 51, 86, 93, 99, 26, 54, 85, 71, 88, 40, 44, 56, 55, 57, 78, 53, 81, 78, 65, 78.
The data have to be grouped to find the mean marks. One of the class intervals made is 60-70.

What is the frequency of the last class interval?

Q. The following are the marks obtained in Maths by students of Class 8:
51, 94, 34, 20, 50, 81, 81, 90, 76, 34, 51, 86, 93, 99, 26, 54, 85, 71, 88, 40, 44, 56, 55, 57, 78, 53, 81, 78, 65, 78.

Group the data and find out how many students scored between 60 and 70?

Q. Convert
the following table into cumulative frequency table.

X	$11$	$12$	$23$	$34$	$56$	$78$
F	$0$	$1$	$2$	$3$	$5$	$6$

Q. The median of the data

78, 56, 22, 34, 45, 54, 39, 68, 54, 84

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The value of 12⋅34⋅56⋅78⋯99100 lies in the interval,

The correct option is A (115,110)Let X=12⋅34⋅56⋅78⋯99100 ⇒X<23⋅45⋅67⋅89⋯100101 as nn+1<n+1n+2 for n>0 ⇒X2<12⋅23⋅34⋅45⋯99100⋅100101=1101<1100 ⇒X<110 Also x><12⋅23⋅34⋅45⋯9899⋅99100=1200<1225 ⇒X<115 ∴115<X<110⇒X∈(115,110)

The value of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \dots \frac{99}{100}$
lies in the interval,