Sum to infinite terms of a gp - class-XI

The series is 3 - 1 + 1/3 - 1/9 + ... which can be split into 3 + (-1 + 1/3 - 1/9 + ...). The part in parentheses is a geometric series with first term a = -1 and common ratio r = -1/3. The sum is a / (1 - r) = -1 / (1 - (-1/3)) = -1 / (4/3) = -3/4. Adding the initial 3 gives 3 - 3/4 = 9/4.

Above in an infinite A.G.S. with $a = 1, d = 1$ for A.P., $b = \dfrac {1}{3}, r = \dfrac {1}{3}$ for G.P.
$\therefore S _{\infty} = \dfrac {ab}{1 - r} + \dfrac {dbr}{(1 - r)^{2}} = \dfrac {\dfrac {1}{3}}{1 - \dfrac {1}{3}} + \dfrac {1 . \dfrac {1}{3} . \dfrac {1}{3}}{\left (1 - \dfrac {1}{3}\right )^{2}} = \dfrac {1}{2} + \dfrac {1}{4} = \dfrac {3}{4}$
$\therefore P = 3^{S} = 3^{3/4} \therefore P^{1/3} = 3^{1/4}$.

$ar = 2, \dfrac {a}{1 - r} = \dfrac {1}{8}$. Eliminate $r$ and we get
$a^{2} - 8a + 16 = 0 \therefore (a - 4)^{2} = 0$ or $a = 4$

$(9)^{S _{\infty}} = 9^{1/2} = 3\because S _{\infty} = \dfrac {1/3}{1 - (1/3)} = \dfrac {1}{2}$.

${ x }=\dfrac { 1 }{ 1-a } $ ${ y }=\dfrac { 1 }{ 1-b } $ [summing infinite $G.P's$].
$\therefore a=\dfrac { x-1 }{ x } $, $b=\dfrac { y-1 }{ y } $
$\therefore 1+ab+{ a }^{ 2 }{ b }^{ 2 }+...\infty $
$=\dfrac { 1 }{ 1-ab } =\dfrac { 1 }{ 1-\dfrac { (x-1)(y-1) }{ xy }  } =\dfrac { xy }{ x+y-1 }. $

Given x = 1/(1-a) and y = 1/(1-b), we have a = (x-1)/x and b = (y-1)/y. The series 1 + ab + a^2b^2 + ... is a geometric series with sum 1/(1-ab). Substituting a and b: 1 / (1 - ((x-1)/x)((y-1)/y)) = 1 / (1 - (xy - x - y + 1)/(xy)) = xy / (xy - xy + x + y - 1) = xy / (x + y - 1).

The series y = x - x^2 + x^3 - x^4 + ... is a geometric series with first term a = x and common ratio r = -x. The sum is y = x / (1 - (-x)) = x / (1 + x). Solving for x: y(1 + x) = x, so y + xy = x, which means y = x - xy = x(1 - y). Thus, x = y / (1 - y).

We can rewrite the series as
$1+1+\dfrac {5}{x}+(\dfrac {5}{x})^2+(\dfrac {5}{x})^3+.....$
We can sum up this series if $\mid 5/x\mid < 1$
$\Leftrightarrow \mid x\mid > 5$

This is the same identity derived in question 466317. The sum of the series 1 + ab + a^2b^2 + ... is indeed xy / (x + y - 1) given the definitions of x and y.

$x^{1/2}.x^{1/4}......\infty =x^{\left ( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} \right )}$ 

 $(8)^{1+|cos x|+|cos x|^2+|cos x|^3+...)}=4^3$
$\Rightarrow (8)^{\dfrac{1}{1-|\cos x|}}=4^3=64=8^2$, since $|\cos x| < 1$ in $(-\pi, \pi)$
$\Rightarrow {\dfrac{1}{1-|\cos x|}}=2$
$\Rightarrow |\cos x|=\cfrac{1}{2}$
The solution in the given interval is,
$x=\pm \cfrac{\pi}{3}, \pm \cfrac{2\pi}{3}$
Hence, option 'C' is correct.

$7+1+\dfrac 17 ...$

The series $4,2,1,\cdots$ form GP

$y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $
 $ =x^{\cfrac{1}{3}+\cfrac{1}{3^2}+\cfrac{1}{3^3}+........\infty }=x^{\cfrac{1/3}{1-1/3}}=x^{1/2}$
Hence, option 'C' is correct.

We know that sum of infinite geometric series $=\dfrac{a}{1-r}$
Where $a=\text{first term}$ and $r=\text{common ratio}$.
$\dfrac{a}{1-r}=\dfrac{4}{3}$
Then, $\dfrac{\dfrac{3}{4}}{1-r}=\dfrac{4}{3}\Rightarrow,r=1-\dfrac{9}{16}=\dfrac{7}{16}$

The series $1-x+x^2-....$ form $GP$ with $a=1 ,r=-x$

Given $log _{2}x+log _{2}(\sqrt{x})+log _{2} (\sqrt[4]{x})+log _{2}(\sqrt[8]{x})+...\infty =4 $

For the given series

$S _{n}=\dfrac{a(1-r^n)}{(1-r)}$

$f\left( x \right) =\dfrac { x }{ 1+x } \ x=\dfrac { f^{ -1 }\left( x \right)  }{ 1-f^{ 1 }\left( x \right)  } \ \therefore f^{ 1 }\left( x \right) =\dfrac { x }{ 1-x } $

${ 9 }^{ \cfrac { 1 }{ 3 }  }\times { 9 }^{ \cfrac { 1 }{ 9 }  }\times { 9 }^{ \cfrac { 1 }{ 27 }  }\times ...\infty $

$x=\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 16 } +....,a=0.2,b\sqrt { 5 } \ x\cfrac { \cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 2 }  } \ x=\cfrac { 1 }{ 2 } \ { a }^{ \log _{ b }{ x }  }=(0.2)^{ \log _{ \sqrt { 5 }  }{ \cfrac { 1 }{ 2 }  }  }\ =(\cfrac { 1 }{ 2 } )^{ \log _{ \sqrt { 5 }  }{ 0.2 }  }\ =(\cfrac { 1 }{ 2 } )^{ -2 }\ a^{ \log _{ b }{ x }  }=4$

$0<x,y,a,b<1\ S=\sqrt { x } (\sqrt { a } +\sqrt { x } )+\sqrt { x } (\sqrt { ab } +\sqrt { xy } )+\sqrt { x } (b\sqrt { a } +y\sqrt { x } )\ =\sqrt { x } \left[ (\sqrt { a } +\sqrt { ab } +b\sqrt { a } ....) \right] +(\sqrt { x } +\sqrt { xy } +y\sqrt { x } )\ =\sqrt { x } \left[ (\cfrac { \sqrt { a }  }{ 1-\sqrt { b }  } )+\cfrac { \sqrt { x }  }{ 1-\sqrt { y }  }  \right] \ S=\cfrac { \sqrt { ax }  }{ 1-\sqrt { b }  } +\cfrac { x }{ 1-\sqrt { y }  } $

$1,r,r^2,r^3..........,(r<0)$

Let the four terms be $\dfrac { a }{ { r }^{ 2 } } ,\dfrac { a }{ r } ,a,a{ r }$.
Therefore, $\frac { a }{ { r }^{ 2 } } *\frac { a }{ r } aa{ r }=4$
$\Rightarrow \dfrac{a^{ 4 }}{r^2}=4$
Also, given that second term is a reciprocal of the fourth term.
So, $\dfrac{a}{ r }=\dfrac { 1 }{ ar } $
$a=\pm 1, r=\pm \dfrac{1}{2}$
Since the sum of infinite terms of G.P is given by $\frac { a }{ 1-r } $. So, by substituting the values of $a$ and $r$; we get
$\frac { a }{ 1-r } =\pm 2$ or $\pm \dfrac{2}{3}$

given $a+ar+ar^{2}+ar^{3}+.......=15$ 

$\begin{array}{l} x={ 4^{ 1/2 } }{ 4^{ 1/4 } }{ 4^{ 1/8 } }\cdots \infty  \ ={ 4^{ \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\cdots +\infty  } } \ ={ 4^{ \frac { { 1/2 } }{ { 1-\frac { 1 }{ 2 }  } }  } } \ =4 \ { \left( { x-4 } \right) ^{ 2 } }=0 \ { x^{ 2 } }+16-8x=0 \end{array}$

Given series is $3,1,\dfrac 13,..$

Let, ${ S } _{ \infty  }=\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 6 } +d\frac { 1 }{ 18 } +..\infty $

$\Rightarrow { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( 1+\dfrac { 1 }{ 3 } +\dfrac { 1 }{ { 3 }^{ 2 } } +....\infty  \right) $

As we know that, sum of infinite G.P series $=\dfrac { a }{ 1-r } $

Therefore, $ { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ 1-\left( 1/3 \right)  }  \right) =\dfrac { 3 }{ 4 } $

Ans: B

Given $\displaystyle \left | a \right |< 1,\left | b \right |< 1,\left | c \right |< 1,\ \ a,b,c \in A.P$

${ 2 }^{ \left( 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty  \right)  }={ 2 }^{ 2 }\ \Rightarrow 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty =2\ \Rightarrow \dfrac { 1 }{ 1-\left| \cos { x }  \right|  } =2\ \Rightarrow 1-\left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow \left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow x=\dfrac { \pi  }{ 3 } ,\dfrac { 2\pi  }{ 3 } $
  in the range $\left( 0,\pi  \right) $

This is an arithmetico-geometric series of the form sum(n^2 * r^(n-1)). The sum can be found using the method of differences or differentiation of geometric series. The result for this specific series is 1/2.

Let common ratio is $r<1$
Then G.P is $1,r,{ r }^{ 2 },{ r }^{ 3 },...\infty $
$S=1+r+{ r }^{ 2 }+{ r }^{ 3 }+...\infty $
$\displaystyle \Rightarrow S=\frac { 1 }{ 1-r } $
Then G.P formed by squaring the terms 
$1,{ r }^{ 2 },{ r }^{ 4 },{ r }^{ 6 },...\infty $
$\displaystyle { S }'=\frac { 1 }{ 1-{ r }^{ 2 } } =\frac { 1 }{ \left( 1-r \right) \left( 1+r \right)  } =\frac { { S }^{ 2 } }{ 2S-1. } $

The series x converges to a value $\dfrac{1}{1-{\cos^2 \phi}};=;{cosec^2 \phi}$

Similarly y converges to  a value ${sec^2 \phi} $

In a similar fashion.. z converges to $\dfrac{{cosec^2 \phi}{sec^2 \phi}}{{cosec^2 \phi}{sec^2 \phi} - 1}$
 $ z = \dfrac{xy}{xy - 1} $

$xy + z = xyz&gt;$; Option B.
And, also
$x + y = xy$
$\therefore xyz = x+ y+ z&gt;$; Option C.

Given that first term $a=-1$ and common ratio is $r=\dfrac{1}{2}$

Sum of a GP $a,ar,ar^2......$

Clearly every summation is infinite series,
$a=\cfrac{1}{1-x}, b = \cfrac{1}{1-y}$ and $c=\cfrac{1}{1-xy}$
or $x=1-\cfrac{1}{a}, y=1-\cfrac{1}{b}$
Simplifying above equation we get, $ac+bc=ab+c$

Let $\displaystyle S=\frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ to $\infty$   ...(1)

Let r be the common ration.
GIven, $S _\infty=S=\frac{a}{1-r}$
$\Rightarrow 1-r=\frac aS$
$\Rightarrow r=1-\frac aS$
Now, sum of n terms is given by
$S _n=\dfrac{a(1-r^n)}{1-r}$

The given series is a Geometric Progression, with first terms $ a =2$ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {1}{2} $

For a GP, sum to infinity is given by the formula $ \dfrac {a}{1-r} $

So, for the given series, $ S _\infty  = \dfrac {2}{1-\dfrac {1}{2}} = 4 $

Given series is $1,-\dfrac{1}{3},+\dfrac{1}{9},-\dfrac{1}{27}......................$

Given the geometric series is $2,\left ( -\dfrac{1}{2} \right ),\left ( \dfrac{1}{8} \right ),\left ( -\dfrac{1}{32} \right ).......................$

Sum of GP $=\dfrac{a(r^n -1)}{r-1}$

$\begin{array}{l} { S _{ p } }, , is, , m-1+{ r^{ p } }+{ r^{ 2p } }... \ { X _{ p } }=1-{ r^{ p } }+{ r^{ 2p } } \ { S _{ p } }=\frac { 9 }{ { 1-R } } =\frac { 1 }{ { 1-{ r^{ p } } } }  \ { X _{ p } }=\frac { 1 }{ { 1+{ r^{ p } } } }  \ { S _{ p } }+{ X _{ p } }=\frac { 1 }{ { 1-{ r^{ p } } } } +\frac { 1 }{ { 1+{ r^{ p } } } }  \ =\frac { { 1+{ r^{ p } }+1 } }{ { \left( { 1-{ r^{ p } } } \right) \left( { 1+{ r^{ p } } } \right)  } } =\frac { 2 }{ { 1-{ r^{ 2p } } } }  \ =2{ S _{ p } } \  \end{array}$

Lets say the sum of infinite geometric series is:

${ S } _{ 1 }=$Sum of n terms in G.P.

$S=\left( 1-\cfrac { 1 }{ x }  \right) +{ \left( 1-\cfrac { 1 }{ x }  \right)  }^{ 2 }+{ \left( 1-\cfrac { 1 }{ x }  \right)  }^{ 3 }+........$
This is a G.P whose first term, $a=1-\cfrac { 1 }{ x } $ and common ration,  $r=1-\cfrac { 1 }{ x } $
${ S } _{ \infty  }=\cfrac { a }{ 1-r } $
$=\cfrac { 1-\cfrac { 1 }{ x }  }{ 1-\left( 1-\cfrac { 1 }{ x }  \right)  } $
$=\cfrac { { \left( x-1 \right)  }/{ x } }{ { \left[ x-\left( x-1 \right)  \right]  }/{ x } } $
$=\cfrac { x-1 }{ x-x+1 } $
$=x-1$

Consider, $y=exp\left[ \left( \sin ^{ 2 } x+\sin ^{ 4 } x+\sin ^{ 6 } x+....+\infty  \right) \log _{ e } 2 \right] $
$\displaystyle\Rightarrow y=exp\left[ \left( \frac { \sin ^{ 2 }{ x }  }{ 1-\sin ^{ 2 }{ x }  }  \right) \log _{ e } 2 \right] =exp\left[ \tan ^{ 2 }{ x } \log _{ e } 2 \right] ={ 2 }^{ \tan ^{ 2 }{ x }  }$
Since, $y$ satisfies $x^{ 2 }-9x+8=0$, then
        $y=1,8$
$\Rightarrow { 2 }^{ \tan ^{ 2 }{ x }  }={ 2 }^{ 0 },{ 2 }^{ 3 }$
$\Rightarrow \tan ^{ 2 }{ x } =0,3$
$\Rightarrow \tan { x } =0,\pm \sqrt { 3 } $

Now, $\displaystyle g\left( x \right) =\frac { \cos  x }{ \cos  x+\sin  x } =\frac { 1 }{ 1+\tan { x }  } =1,\frac { 1 }{ 1\pm \sqrt { 3 }  } =1,\frac { -\sqrt { 3 } -1 }{ 2 } ,\frac { \sqrt { 3 } -1 }{ 2 } $

Ans: B

$\displaystyle e^{(\cos^{2}x+\cos^{4}x+\cdots )\log{3}}=e^{\left(\dfrac{\cos^{2}x}{1-\cos^{2}x}\right)\log3.}$
$\because (\cos^{2}x+\cos^{4}x+\cdots)$ is forming an infinite G.P. $=e^{\cot^{2}x\log3.}$
$y^{2}-10y+9=0  \Rightarrow   y=9,1$
$e^{\cot^{2}x\log{3}}=9,1$
$3^{\cot^{2}x}=9,1&gt;&gt;[\because e ^{\log x}=x]$
$\Rightarrow \cot^{2}x=2,0$ but as $x\in \left[0,\dfrac{\pi }{2}\right]$
$\Rightarrow \cot^{2}x=0$

$x+y=a$