Tag: sum to infinite terms of a gp

Questions Related to sum to infinite terms of a gp

If the sum of an infinite $G.P.$ is $1$ and the second term is $'x'$.

Then the range of $'x'$ is

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

  1. $\left( 0,\dfrac { 1 }{ 4 } \right]$

  2. $\left[ -2,\dfrac { 1 }{ 4 } \right]$

  3. $(-2, 0)$

  4. $[-2, 0]$

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Correct Option: A

The value of $a^{\log _{2}}x$, where $a=0.2,b=\sqrt {5},x=\dfrac {1}{4}+\dfrac {1}{8}+\dfrac {1}{16}+.....$ to $\infty $ is

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

  1. $1$

  2. $2$

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

  4. $4$

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Correct Option: D
Explanation:

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

If $0<x,y,a,b<1$,then the sum of infinite terms of the series $\sqrt x (\sqrt a  + \sqrt x ) + \sqrt x (\sqrt {ab}  + \sqrt {xy} ) + \sqrt x (b\sqrt a  + y\sqrt x ) + .......$ is

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

  1. $\dfrac{{\sqrt {ax} }}{{1 + \sqrt b }} + \dfrac{x}{{1 + \sqrt y }}$

  2. $\dfrac{{\sqrt x }}{{1 + \sqrt b }} + \dfrac{{\sqrt x }}{{1 + \sqrt y }}$

  3. $\dfrac{{\sqrt x }}{{1 - \sqrt b }} + \dfrac{{\sqrt x }}{{1 - \sqrt y }}$

  4. $\dfrac{{\sqrt {ax} }}{{1 - \sqrt b }} + \dfrac{x}{{1 - \sqrt y }}$

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Correct Option: D
Explanation:

$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 }  } $

If $A = 1 + {r^a} + {r^{2a}} + {r^{3a}}......\infty $ and $B = 1 + {r^b} + {r^{2b}}......\infty$ then$\dfrac{a}{b} = $

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

  1. $\dfrac{\log{\left({A-1}\right)}}{\log{\left({B-1}\right)}}$

  2. $\dfrac{\log{\left(\dfrac{A-1}{A}\right)}}{\log{\left(\dfrac{B-1}{B}\right)}}$

  3. $\dfrac{\log{\left({A}\right)}}{\log{\left({B}\right)}}$

  4. $\dfrac{\log{\left({B}\right)}}{\log{\left({A}\right)}}$

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Correct Option: B
Explanation:

If $\left|Common\, ratio\right|<1$, then sum of infinite terms of G.P. is

$S=\dfrac{First\,  Term}{1\,−\,Common \,Ratio}$

$A=1+{r}^{a}+{r}^{2a}+{r}^{3a}+...$

Common ratio$=\dfrac{{r}^{a}}{1}={r}^{a}$

$A=\dfrac{1}{1−{r}^{a}}$

$\Rightarrow\,1−{r}^{a}=\dfrac{1}{A}$

$\Rightarrow\,{r}^{a}=1−\dfrac{1}{A}$

$\Rightarrow\,{r}^{a}=\dfrac{A−1}{A}$

Take $\log$ on both

$\log{\left({r}^{a}\right)}=\log{\left(\dfrac{A−1}{A}\right)}$

As $\log{\left({m}^{n}\right)}=n\log{\left(m\right)}$

$\Rightarrow\,a\log{\left(r\right)}=\log{\left(\dfrac{A−1}{A}\right)}$  ....$(i)$

$B=1+{r}^{b}+{r}^{2b}+{r}^{3b}+...$

Common ratio$=\dfrac{{r}^{b}}{1}={r}^{b}$

$B=\dfrac{1}{1−{r}^{b}}$

$\Rightarrow\,1−{r}^{b}=\dfrac{1}{B}$

$\Rightarrow\,{r}^{b}=1−\dfrac{1}{B}$

$\Rightarrow\,{r}^{b}=\dfrac{B−1}{B}$

Take $\log$ on both

$\log{\left({r}^{b}\right)}=\log{\left(\dfrac{B−1}{B}\right)}$

As $\log{\left({m}^{n}\right)}=n\log{\left(m\right)}$

$\Rightarrow\,b\log{\left(r\right)}=\log{\left(\dfrac{B−1}{B}\right)}$  ....$(ii)$

$\dfrac{(i)}{(ii)}=\dfrac{a\log{r}}{b\log{r}}=\dfrac{\log{\left(\dfrac{A-1}{A}\right)}}{\log{\left(\dfrac{B-1}{B}\right)}}$

$\therefore\,\dfrac{a}{b}=\dfrac{\log{\left(\dfrac{A-1}{A}\right)}}{\log{\left(\dfrac{B-1}{B}\right)}}$

The sum of the terms of an infinitely decreasing G.P. is $S$. The sum of the squares of the terms of the progression is -

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

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

  2. $\dfrac{{{S^2}}}{{2S - 1}}$

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

  4. ${S^2}$

Show answer
Correct Option: B
Explanation:

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


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

$S=\dfrac {1}{1-r}$


$r=1-\dfrac {1}{S}=\dfrac {S-1}{S}$

$S _{\infty}=1^2+r^2+r^4+r^6+..........$

$S _{\infty}=\dfrac {1}{1-r^2}$

       $=\dfrac {1}{1- \left (\dfrac {S-1}{S}\right )^2}$

       $=\dfrac {S^2}{S^2-S^2-1+2S}$

        $=\dfrac {S^2}{2S-1}$

In a GP the product of the first four terms is 4 and the second term is the reciprocal of the fourth term. The sum of the GP up to infinite terms is-

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

  1. $2$

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

  3. $-2$

  4. 6

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Correct Option: A,B,C
Explanation:

Let the four terms be $\dfrac { a }{ { r }^{ 2 } } ,\dfrac { a }{ r } ,a,a{ r }$.
Therefore, $\frac { a }{ { r }^{ 2 } } *\frac { a }{ r } *a*a{ r }=4$
$\Rightarrow \dfrac{a^{ 4 }}{r^2}=4$
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 $\pm \dfrac{2}{3}$

Sum to infinity of a G.P is $15$, whose first term is $a$ then a MUST satisfy the inequality given by

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

  1. $0< a< 130$

  2. $0< a< 30$

  3. $0< a< 15$

  4. $0< a< 100$

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Correct Option: C
Explanation:

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

where $0 < r < 1  \dfrac{a}{1-r}=15$ 
$(\because |r|\geqslant 1$, geometric series is divergent $)$
$a=15(1-r)$
when $0 < r < 1 $
$\Rightarrow -1 < -r < 0.$
$\Rightarrow  0 < 1-r < 1$
$\therefore  0 < 15(1-r) < 15$
$\Rightarrow 0 < a < 15$

If $x=\sqrt{4}.\sqrt[4]{4}. \sqrt[8]{4}.\sqrt[16]{4}........ \infty$, then 

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

  1. $x^2-8x+16=0$

  2. $x^2-3x+2=0$

  3. $x^2-5x+4=0$

  4. $x^2+5x+4=0$

Show answer
Correct Option: A
Explanation:

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

The sum of  $3,1,\dfrac 13 ,....$ is

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

  1. $\dfrac 52$

  2. $\dfrac 92$

  3. $\dfrac 72$

  4. $\dfrac {11}2$

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Correct Option: B
Explanation:

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


Given series is in GP.

The common ratio is given as $\dfrac{1}{3}$

The sum of infinite terms is  $\dfrac{a}{1-r}$

$\implies  \dfrac 3{1-\dfrac 13}$

$\implies \dfrac{3}{\dfrac 23}=\dfrac 92$


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

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

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

  2. $\dfrac{1}{4}$

  3. $\dfrac{3}{5}$

  4. $1$

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Correct Option: C
Explanation:
Let $a,ar,a{r}^{2},...$ to $\infty$
Now, sum of infinite G.P$=20$
$\Rightarrow \dfrac{a}{1-r}=20$          .........$\left(1\right)$
where each term of the above G.P is squared, then the progression becomes
${a}^{2},{a}^{2}{r}^{2},{a}^{2}{r}^{4},.....$
Now, first term $A={a}^{2}$ and common ratio$R={r}^{2}$
Sum of above G.P$=100$
$\Rightarrow \dfrac{A}{1-R}=100$
$\Rightarrow \dfrac{{a}^{2}}{1-{r}^{2}}=100$     ....$\left(2\right)$
Squaring $\left(1\right)$ we get
$\dfrac{{a}^{2}}{{\left(1-r\right)}^{2}}=400$    ....$\left(3\right)$
Dividing eqn$\left(2\right)$ by $\left(3\right)$ we get
$\dfrac{{\left(1-r\right)}^{2}}{1-{r}^{2}}=\dfrac{400}{100}=4$
$\Rightarrow \dfrac{{\left(1-r\right)}^{2}}{\left(1-r\right)\left(1+r\right)}=4$
$\Rightarrow \dfrac{1-r}{1+r}=4$
$\Rightarrow 4-4r=1+r$
$\Rightarrow 5r=3$
$\therefore r=\dfrac{3}{5}$