Tag: sum to infinite terms of a gp

Questions Related to sum to infinite terms of a gp

If $S _{1}, S _{2}, S _{3}$ are respectively the sum of n, 2n and 3n terms of a G.P. Then  $S _{1}(S _{3}-S _{2}) = (S _{2} -S _{1})^{2}$.

sum to infinite terms of a gp sequence, progression and series maths

  1. True

  2. False

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Correct Option: A
Explanation:

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

${ S } _{ 2 }=$Sum of 2n terms in G.P.
${ S } _{ 3 }=$4Sum of 3n terms in G.P.
${ S } _{ 1 }=\cfrac { a({ r }^{ n }-1) }{ r-1 } ,{ S } _{ 2 }=\cfrac { a({ r }^{ 2n }-1) }{ r-1 } ,{ S } _{ 3 }=\cfrac { a({ r }^{ 3n }-1) }{ r-1 } $
a is a first term and r is a common ratio.
${ S } _{ 3 }-{ S } _{ 2 }=\cfrac { a({ r }^{ 3n }-{ r }^{ 2n }) }{ r-1 } \ { S } _{ 1 }({ S } _{ 3 }-{ S } _{ 2 })=\cfrac { { a }^{ 2 }({ r }^{ 4n }-2{ r }^{ 3n }+{ r }^{ 2n }) }{ { (r-1 })^{ 2 } } \ =\cfrac { { a }^{ 2 }({ r }^{ 2n }-{ r }^{ n })^{ 2 } }{ { (r-1 })^{ 2 } } \ =\cfrac { { a }^{ 2 }(({ r }^{ 2n }-{ 1)(r }^{ n }-1))^{ 2 } }{ { (r-1 })^{ 2 } } \ =\left[ \cfrac { { a }({ r }^{ 2n }-{ 1)-a(r }^{ n }-1) }{ { (r-1 }) }  \right] ^{ 2 }\ { S } _{ 1 }({ S } _{ 3 }-{ S } _{ 2 })=({ S } _{ 2 }-{ S } _{ 1 })^{ 2 }$

If $|x| > 1$, then
$\left(1-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{x}\right)^2+\left(1-\dfrac{1}{x}\right)^3+.....=$

sum to infinite terms of a gp sequence, progression and series maths

  1. $x-1$

  2. $x+1$

  3. $x$

  4. $\dfrac{1}{x-1}$

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Correct Option: A
Explanation:

$S=\left( 1-\cfrac { 1 }{ x }  \right) +{ \left( 1-\cfrac { 1 }{ x }  \right)  }^{ 2 }+{ \left( 1-\cfrac { 1 }{ x }  \right)  }^{ 3 }+........$
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 } $
$=x-1$

If $e^{\displaystyle \left [ \left ( \sin^{2}x + \sin^{4}x + \sin^{6}x + .... + \infty \right ) \log _{e}2\right ]}$ satisfies the equation $\displaystyle x^{2} -9x + 8 = 0$,then the value of $\displaystyle g \left ( x \right ) = \frac{\cos x}{\cos x + \sin x}$ is

sum to infinite terms of a gp sequence, progression and series maths

  1. $\displaystyle \frac{\sqrt{3} + 1}{2}$

  2. $\displaystyle \frac{\sqrt{3} - 1}{2}$

  3. $\displaystyle 8$

  4. None of these

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Correct Option: B
Explanation:

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

lf $e^{(\cos^{2}x+\cos^{4}x+\cos^{6}x+\ldots.)\log 3}$ satisfies $y^{ 2 }-10y+9=0$ and $0\le x\le \cfrac { \pi  }{ 2 } $, then $\cot^{2}x=$

sum to infinite terms of a gp sequence, progression and series maths

  1. $0$

  2. $1$

  3. $\dfrac12$

  4. $9$

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Correct Option: A
Explanation:

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

Hence, option 'A' is correct.

If the sum of an infinite $GP$ is $20$ and sum of their square is $100$ then common ration will be=

sum to infinite terms of a gp sequence, progression and series maths

  1. $1/2$

  2. $1/4$

  3. $3/5$

  4. $1$

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Correct Option: C
Explanation:
Let $G.P$ is $a, ar, ar^2,....\infty$
sum $=\dfrac {a}{1-r}=20---(1)$
If terms are required the $G.P.$
becomes : $a^2, a^2r^2, a^2 r^4,.....\infty$
Sum $=\dfrac {a^2}{1-r^2}=100---(2)$
Name dividing square of equation $(1)$ by equation $(2)$
$\dfrac {\dfrac {a^2}{1-r^2}}{\dfrac {a^2}{(1-r)^2}}=\dfrac {100}{400}$
$\Rightarrow \ \dfrac {(1-r)}{(1-r)(1+r)}=\dfrac {1}{4}$
$\Rightarrow \ \dfrac {1-r}{1+r}=\dfrac {1}{4}$
$\Rightarrow \ 4-4r=1+r$
$\Rightarrow \ 4-1=4r+r$
$\Rightarrow \ 5r=3$
$\Rightarrow \ r=\dfrac {3}{5}$

For $0 < \phi < \pi/2$ if $x=\sum _{n=0}^{\infty }\cos ^{2n} \phi, y=\sum _{n=0}^{\infty }\sin ^{2n} \phi, z=\sum _{n=0}^{\infty }\cos ^{2n} \phi \sin^{2n}\phi$, then 

sum to infinite terms of a gp sequence, progression and series maths

  1. $xyz=xz+y$

  2. $xyz=xy+z$

  3. $xyz=x+y+z$

  4. $xyz=yz+x$

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Correct Option: B
Explanation:

Given $\displaystyle x=\sum _{ n=0 }^{ \infty  }{ { \cos }^{ 2n }\phi ,\quad y=\sum _{ n=0 }^{ \infty  }{ { \sin }^{ 2n }\phi  }  } $ and $\displaystyle z=\sum _{ n=0 }^{ \infty  }{ { \cos }^{ 2n }\phi  } { \sin }^{ 2n }\phi $


\since $\displaystyle 0<\phi <\frac { \pi  }{ 2 } $, so each series is geometric series with common ratio $\displaystyle r<1$. Therefore, the series are convergent.


Now, $\displaystyle x=\frac { 1 }{ 1-{ \cos }^{ 2 }\phi  } $


$\displaystyle =\frac { 1 }{ { \sin }^{ 2 }\phi  } $      $(\because S _{ \infty  }=\dfrac { a }{ 1-r } )$


$\displaystyle y=\frac { 1 }{ 1-{ \sin }^{ 2 }\phi  } $      $(\because S _{ \infty  }=\dfrac { a }{ 1-r } )$


$\displaystyle =\frac { 1 }{ { \cos }^{ 2 }\phi  } $


$\displaystyle z=\frac { 1 }{ 1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi  } $ 

   

$(\because S _{ \infty  }=\dfrac { a }{ 1-r } )$


Consider, $\displaystyle xyz=\frac { 1 }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi (1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi ) } $      $(1)$


Also, $\displaystyle =\frac { 1 }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi  } +\frac { 1 }{ 1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi  } $


$\displaystyle xy+z=\frac { 1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi +{ \sin }^{ 2 }\phi { \cos }^{ 2 }\phi  }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi (1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi ) } $


$\displaystyle =\frac { 1 }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi (1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi ) } $

 

$\displaystyle =xyz$      $[From (1)]$

 

The sum of the intercepts cut off by the axes on the lines  $ x+y=a,x+y=ar,x+y=ar^{2}\ldots\ldots\ldots$ where $a\neq 0$ and $r=\displaystyle \dfrac{1}{2}$  is 

sum to infinite terms of a gp sequence, progression and series maths

  1. $2a$

  2. $a\sqrt{2}$

  3. $2\sqrt{2}a$

  4. $ \displaystyle \dfrac{a}{\sqrt{2}}$

Show answer
Correct Option: C
Explanation:

$x+y=a$

intercept cut off by the axes $=\sqrt{a^2+a^2}=\sqrt{2}a$

$\therefore$ sum of all intercepts cut off by the axes on the lines,

$x+y=a, x+y=ar,.... x+y=a^{r^n-1}$

$\sqrt{2}a, \sqrt{2}ar,.... \sqrt{2}ar^{n-1}$

$Sum=\sqrt{2}a+\sqrt{2}ar+....+\sqrt{2}a^{r^n-1}....$as

$=\sqrt{2}(\dfrac{a}{1-r})$

$=\dfrac{\sqrt{2}a}{1-r}$

Given  $\Rightarrow r=\dfrac{1}{2}$

$\therefore Sum=2\sqrt{2}a$