Tag: maths

Questions Related to maths

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. $\dfrac {108}{9}$

Show answer
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. $81$

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

Show answer
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 $^an = \dfrac{a^n \left(1-\left(\dfrac{b}{a}\right)^{n+1}\right)}{1-b/a} = \dfrac{a^n(a^{n+1} - b^{n+1} )/a^{n+1}}{\dfrac{(a-b)}{a}}$
$= \dfrac{a^{n+1}-b^{n+1}}{a-b}$
$= \dfrac{b^{n+1}-a^{n+1}}{b-a}$

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. $[-2, 0]$

Show answer
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. $4$

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

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

Show answer
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

Show answer
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 $\pm \dfrac{2}{3}$