Tag: maths

Questions Related to maths

Find the sum of $4,2,1,\cdots$ 

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

  1. 8

  2. 16

  3. 32

  4. 64

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Correct Option: A
Explanation:

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

With $a=4$ and $r=\dfrac 12$
$S _{\infty}=\dfrac{a}{1-r}\\dfrac{4}{1-\dfrac 12 }=\dfrac{4}{\dfrac 12}=4\times 2=8$

If $y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $, then $y =$

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

  1. $x^{1/3}$

  2. $x^{2/3}$

  3. $x^{1/2}$

  4. $x$

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Correct Option: C
Explanation:

$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.

If sum of an infinite geometric series is $\dfrac{4}{3}$ and its Ist term is $\dfrac{3}{4}$, then its common ratio is

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

  1. $\dfrac{7}{16}$

  2. $\dfrac{9}{16}$

  3. $\dfrac{1}{9}$

  4. $\dfrac{7}{9}$

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Correct Option: A
Explanation:

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 value of x that satisfies the relation 
$x=1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }-{ x }^{ 5 }+........\infty $ 

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

  1. $2cos{ 3 }6^{ \circ }$

  2. $2cos144^{ \circ }$

  3. $2sin18^{ \circ }$

  4. none

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Correct Option: A
Explanation:

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

Sum of infinte GP is $x=\dfrac{a}{1-r}\x=\dfrac{1}{1+x}\x+x^2=1\x^2+x-1=0$
By quadratic formulae 
$x=\dfrac{-1\pm\sqrt{1+4}}2\2\dfrac{-1\pm\sqrt5}{4}\2\cos 36^{\circ}$

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

Show answer
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 } $

Show answer
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. $2$

Show answer
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

Show answer
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 } $