Tag: maths

By rationalising each term,
$\frac {1}{1+\sqrt 2}=\frac {1}{\sqrt 2+1}\times \frac {\sqrt 2-1}{\sqrt 2-1}=\frac {\sqrt 2-1}{2-1}=\sqrt 2-1$
$\frac

{1}{\sqrt 2+\sqrt 3}=\frac {1}{\sqrt 3+\sqrt 2}\times \frac {\sqrt

3-\sqrt 2}{\sqrt 3-\sqrt 2}=\frac {\sqrt 3-\sqrt 2}{3-2}$
$=\sqrt 3-\sqrt 2$
............................
$\frac {1}{\sqrt {15}+4}=\frac {1}{4+\sqrt {15}}\times \frac {4-\sqrt {15}}{4-\sqrt {15}}=\frac {4-\sqrt {15}}{16-15}$
$=4-\sqrt {15}$
$\therefore$ Given expression
$=(\sqrt 2-1)+(\sqrt 3-\sqrt 2)+.....+4-\sqrt {15}$
$=4-1=3$.

Add the numbers $4, 6, 8, 10, 12$ etc Next number is $31 +12 = 43$

Let $\displaystyle S=1+\frac { 4 }{ 5 } +\frac { 7 }{ { 5 }^{ 2 } } +\frac { 10 }{ { 5 }^{ 3 } } +...$   ...(1)
Multiply (1) by $\displaystyle \frac { 1 }{ 5 } $ , we get

$\displaystyle \frac { 1 }{ 5 } S=\frac { 1 }{ 5 } +\frac { 4 }{ { 5 }^{ 2 } } +\frac { 7 }{ { 5 }^{ 3 } } +\frac { 10 }{ { 5 }^{ 4 } } +...$   ...(2)

$(1) -(2)$, gives 
$\displaystyle \left( 1-\frac { 1 }{ 5 }  \right) S=1+\frac { 3 }{ 5 } +\frac { 3 }{ { 5 }^{ 2 } } +\frac { 3 }{ { 5 }^{ 3 } } +...$

$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 5 } \left( 1+\frac { 1 }{ 5 } +\frac { 1 }{ { 5 }^{ 2 } } +... \right) $

$\displaystyle \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 1 }{ 1-\dfrac { 1 }{ 5 }  }  \right) \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 5 }{ 4 }  \right) $

$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 4 } \Rightarrow S=\frac { 35 }{ 16 } $

$1+4 + 9r^2 + 16r^3 ....\infty$

Given that $a = 2; d = 3$
$n = 25 ; a _{25} = $?
Using Gauss method, we find the value of $a _{25}$
$a _n = a + (n - 1) d$
$a _{25} = 2 + (25 - 1) 3$
$= 2 + (24) 3$
$a _{25} = 74$
Sum of 'n' series using Gauss method is,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$= \dfrac{25}{2} [2 + 74]$
$= 12.5 (76)$
$S _n = 950$

Prime numbers in decreasing number series Next number is 53

On simplification of first two term , we observed that
$\left(\frac{2}{3}\right)$ $\times \left(\frac{3}{4}\right)...$
So numerator is the multiplication from 2 to n-1 and denominator is the multiplication from 3 to n
So making it as factorial of first n and n-1 term we have to multiply 1 to numerator and 1 and 2 to denominator
$2\left(\frac{(n-1)!}{n!}\right)= \frac{2}{n}$

$\begin{array}{l} \overrightarrow { a } =2\left( { { { \log   } _{ 3 } }x,a } \right) ,\overrightarrow { b } =\left( { -3,a{ { \log   } _{ 3 } }x,{ { \log   } _{ 3 } }x } \right)  \ \overrightarrow { a } \overrightarrow { b } =\left| a \right| \left| b \right| \cos  \theta  \ \cos  \theta =\frac { { \overrightarrow { a } .\overrightarrow { b }  } }{ { \left| a \right| \left| b \right|  } }  \ =\frac { { -6+a{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+a.{ { \log   } _{ 3 } }x } }{ { \sqrt { 4+{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+{ a^{ 2 } } } .\sqrt { 9+{ a^{ 2 } }{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } } }  } }  \ for\, \, acute\, \, angle\, \, a>0 \ Option\, \, \, C\, \, is\, \, correct. \end{array}$

$\displaystyle a\times b=a\times c\Rightarrow a\times b-a\times c=0$
$\Rightarrow a\times (b-c)=0$ $\Rightarrow a \parallel (b-c)$
$\Rightarrow b-c = \lambda a\Rightarrow b= c+\lambda a$