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Yogita Ingle 5 years, 4 months ago
first 10 natural number=1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
Mean = 55/10 = 5.5
now, 5th and 6th term as the total number is even .
Median = 5+6/2 = 11/ 2 = 5.5
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Sia ? 5 years, 4 months ago
It is given that,
x = 3 + {tex}\sqrt{8}{/tex}
Now, {tex}\frac{1}{x}{/tex} = {tex}\frac{1}{3+\sqrt{8}}{/tex} = {tex}\frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}{/tex} ( rationalising the expression)
= {tex}\frac{3-\sqrt{8}}{(3)^{2}-(\sqrt{8})^{2}}{/tex}
= {tex}\frac{3-\sqrt{8}}{9-8}{/tex} = 3 - {tex}\sqrt{8}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{x}{/tex} = 3 - {tex}\sqrt{8}{/tex}
We know that, (x + {tex}\frac{1}{x}{/tex})2 = x2 + {tex}\frac{1}{x^2}{/tex} + 2 {tex}\times{/tex} x {tex}\times{/tex} {tex}\frac{1}{x}{/tex}
{tex}\Rightarrow{/tex} (3 + {tex}\sqrt{8}{/tex} + 3 - {tex}\sqrt{8}{/tex})2 = x2 + {tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} (6)2 = x2 + {tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} 36 = x2 + {tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} x2 + {tex}\frac{1}{x^2}{/tex} = 36 - 2 = 34
Hence, x2 + {tex}\frac{1}{x^2}{/tex} = 34
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Sia ? 5 years, 4 months ago
Suppose {tex}\sqrt2{/tex} is a rational number. That is , {tex}\sqrt2{/tex} = {tex}\frac{p}{q}{/tex} for some p{tex}\in{/tex}Z and q {tex}\in{/tex}Z. We can assume the fraction is in lowest fraction, That is p and q shares no common factors.
Then {tex}\sqrt2q=p{/tex}
Squaring both side we get,
{tex}2q^2=p^2{/tex}
So {tex}p^2{/tex} is a multiple of 2,
let's assume {tex}p=2m{/tex}
Then, {tex}2q^2=\left(2m\right)^2{/tex}
{tex}2q^2=4m^2{/tex}
Or {tex}q^2=2m^2{/tex}
So {tex}q^2{/tex} is a multiple of 2,
{tex}\therefore{/tex} q is multiple of 2
Thus p and q shares a common factor.This is contradiction.
{tex}\Rightarrow {/tex}{tex}\sqrt { 2 }{/tex} is an irrational number.
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