Tag: sum to infinite terms of a gp

Questions Related to sum to infinite terms of a gp

If $x>0$ and $\displaystyle log _{2}x+log _{2}(\sqrt{x})+log _{2} (\sqrt[4]{x})+log _{2}(\sqrt[8]{x})+...\infty =4 ,$then $x=$

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

  1. 2

  2. 3

  3. 4

  4. 5

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Correct Option: C
Explanation:

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

$\Rightarrow log _{2}[x.x^{1/2}.x^{1/4}.x^{1/8}...\infty

]=log _{2}[x^{1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...\infty} ]=log _{2}

x^{\dfrac{1}{1-(1/2)}}=log _{2}(x^{2})=4 $

$ \therefore x^{2}=2^{4}=16

 \therefore x=4$

What is the sum of the series $ 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + ....$ equal to ?

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

  1. $\dfrac{1}{2}$

  2. $\dfrac{3}{2}$

  3. $2$

  4. $\dfrac{2}{3}$

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Correct Option: D
Explanation:
$1,\dfrac { -1 }{ 2 } ,\dfrac { 1 }{ 4 } ,\dfrac { -1 }{ 8 } ,..$ is in G.P series. 
So, sum of infinite terms of a G.P is $\dfrac { a }{ 1-r } $, where $a$ is first term $=1$
$r$ is common difference $=-1/2=-0.5$
Thus, $1+\dfrac { -1 }{ 2 } +\dfrac { 1 }{ 4 } +\dfrac { -1 }{ 8 } ,..=$ $\dfrac { 1 }{ 1-\left( \frac { -1 }{ 2 }  \right)  } =\dfrac { 1 }{ \left( \frac { 3 }{ 2 }  \right)  } =\dfrac { 2 }{ 3 } $
Hence, D is correct.

The sum of the series formed by the sequence $3, \sqrt{3}, 1....... $ upto infinity is : 

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

  1. $\frac {3\sqrt{3}(\sqrt{3}+1)}{2}$

  2. $\frac {3\sqrt{3}(\sqrt{3} - 1)}{2}$

  3. $\frac {3(\sqrt{3}+1)}{2}$

  4. $\frac {3(\sqrt{3}-1)}{2}$

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Correct Option: A
Explanation:

For the given series

$\ \cfrac { \sqrt { 3 }  }{ 3 } =\cfrac { 1 }{ \sqrt { 3 }  } \$
So it is a GP with common ratio $\cfrac{1}{\sqrt {3}}$
Formula for sum of infinite terms of GP $ =\cfrac { a }{ 1-r }$,  where
$a$ is first term and r is the common ratio
$\Rightarrow  { S } _{ \infty  }=\cfrac { 3 }{ 1-\cfrac { 1 }{ \sqrt { 3 }  }  }  =\cfrac { 3\sqrt { 3 }  }{ \sqrt { 3 } -1 }$

$=\cfrac { 3\sqrt { 3 } (\sqrt { 3 } +1) }{ (\sqrt { 3 } -1)(\sqrt { 3 } +1) }$ ..... [On rationalizing]

$ =\cfrac { 3\sqrt { 3 } (\sqrt { 3 } +1) }{ 2 } $
Hence, A is correct.

In a Geometric progression with common ratio less than $1$, if $n$ approaches $\infty$ then ${ S } _{ \infty  }$ is

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

  1. $a{ r }^{ 0 }$

  2. $a{ r }^{ n-1 }$

  3. $\cfrac { 1-r }{ a } $

  4. $\cfrac { a }{ 1-r } $

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Correct Option: D
Explanation:

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

Now, as $n$ tends to $\infty$ then $r^n$ tends to $0$
$\therefore$ $S _{\infty}=\dfrac{a(1-r^{\infty})}{(1-r)}$
           $=\dfrac{a}{1-r}$       $(\because\displaystyle \lim _{n\rightarrow \infty}r^{\infty} = 0$ for $r<1)$
Hence, $S _{\infty}=\dfrac{a}{(1-r)}$

Find the sum of the infinite geometric series $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.......$

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

  1. $16$

  2. $14$

  3. $-11$

  4. $2$

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Correct Option: D
Explanation:
Given sequence is $1+\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}+....$
Thus $a=1$, $r = \dfrac{1}{2}$
Therefore, $\text{sum} =\dfrac{a}{1-r}$
$\Rightarrow \text{sum} = \dfrac{1}{1-\frac{1}{2}}$
$\Rightarrow \text{sum} =2$

If $p$ is positive, then the sum to infinity of the series, ${1 \over {1 + p}} - {{1 - p} \over {{{(1 + p)}^2}}} + {{{{(1 - p)}^2}} \over {{{(1 + p)}^3}}} - ......$ is

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

  1. $1/2$

  2. $3/4$

  3. $1$

  4. None of these

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Correct Option: A
Explanation:
$a=\dfrac { 1 }{ 1+P } \\ r=-\dfrac { 1-P }{ 1+P } $
Sum to infinity $=\dfrac { a }{ 1-r } \\ =\dfrac { \dfrac { 1 }{ 1+P }  }{ 1+\dfrac { 1-P }{ 1+P }  } \\ =\dfrac { \dfrac { 1 }{ 1+P }  }{ \dfrac { 1+P+1-P }{ 1+P }  } =\dfrac { 1 }{ 2 } $

If $f(x) = x - {x^2} + {x^3} - {x^4} + .............\infty $ where $\left| x \right|\langle 1$ then ${f^{ - 1}}(x) = $

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

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

  2. ${\dfrac{x}{1 + x}}$

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

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

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Correct Option: A
Explanation:

$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 } $

If the sum of an infinitely decreasing G.P. is $3$, and the sum of the squares of its terms is $\dfrac {9}{2}$, then the sum of the cubes of the terms is

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

  1. $\dfrac {105}{13}$

  2. $\dfrac {108}{13}$

  3. $\dfrac {729}{8}$

  4. $\dfrac {108}{9}$

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Correct Option: B
Explanation:
Let the GP be $a,ar,ar^2,ar^3,...$
The first term be $a$ and the common ratio be $r$. 
Then, it is given that:
$\text{Sum}=\dfrac{a}{1-r} = 3$    ....(1)
Sequence of squares of terms is $a^2, a^2r^2, a^2r^4,...$
$\text{Sum}=\dfrac{a^2}{1-r^2} = \dfrac{9}{2}$    ....(2)
Thus, we have:
$\dfrac{a^2\div a}{(1-r^2)\div (1-r)} = \dfrac{9 \div 3}{2} $
$\Rightarrow \dfrac{a}{1+r} = \dfrac{3}{2}$    ....(3)

Now divide equations (1) and (3),
$\dfrac{1-r}{1+r} = \dfrac{1}{2}$
$ \Rightarrow r = \dfrac{1}{3}$
Substituting this value of $r$ in equation (1), we get $a = 2$
Therefore, $\dfrac{a^3}{1-r^3} = \dfrac{8}{\left (1-\dfrac{1}{27}\right)} = \dfrac{108}{13}$
Thus, the answer is option B.

Sum of the series ${9^{{1 \over 3}}} \times {9^{{1 \over 9}}} \times {9^{{1 \over {27}}}} \times .......$  is equal to

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

  1. $3$

  2. $9$

  3. $27$

  4. $81$

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Correct Option: A
Explanation:

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

$={ 9 }^{ \cfrac { 1 }{ 3 }  +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 27 } + ...}$
Let $S=\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 27 } +...$
$=\cfrac { \cfrac { 1 }{ 3 }  }{ 1-\cfrac { 1 }{ 3 }  } $
$=\cfrac { \cfrac { 1 }{ 3 }  }{ \cfrac { 2 }{ 3 }  } =\cfrac { 1 }{ 2 } $
$\therefore { 9 }^{ \cfrac { 1 }{ 3 }  }\times { 9 }^{ \cfrac { 1 }{ 9 }  }\times { 9 }^{ \cfrac { 1 }{ 27 }  }\times ...\infty ={ 9 }^{ \cfrac { 1 }{ 2 }  }$
$=(3^{ 2 })^{ \cfrac { 1 }{ 2 }  }$
$=3$

If the expansion in powers of x of the function $\dfrac{1}{(1 - ax)(1 - bx)} , (a \neq b)$ is $a _0 + a _1x + a _2x^2 + .... \, then \, a _n$ is

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

  1. $\dfrac{b^n - a^n}{b - a}$

  2. $\dfrac{a^n - b^n}{b - a}$

  3. $\dfrac{a^{n+1} - b^{n+1}}{b - a}$

  4. $\dfrac{b^{n+1} - a^{n+1}}{b - a}$

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Correct Option: D
Explanation:
We know $\dfrac{1}{1-ax} = \displaystyle \sum _{r=0}^{\infty} (ax)^r$ 
$1+ax+a^2 x^2+............$
$\dfrac {1}{1-bx} = 1+bx+b^2 x^2 +..........$
So $\dfrac{1}{(1-ax)} \times \dfrac{1}{(1-bx)} = \displaystyle \sum _{r=0}^{\infty} (ax)^r \times \displaystyle \sum _{r=0}^{\infty}(bx)^r$
coefficient of $x^n$ is given by $= a^n + a^{n-1} b + .... + b^n$
given term is $g.p$ will ratio of $b/a$
So $^an = \dfrac{a^n \left(1-\left(\dfrac{b}{a}\right)^{n+1}\right)}{1-b/a} = \dfrac{a^n(a^{n+1} - b^{n+1} )/a^{n+1}}{\dfrac{(a-b)}{a}}$
$= \dfrac{a^{n+1}-b^{n+1}}{a-b}$
$= \dfrac{b^{n+1}-a^{n+1}}{b-a}$