Tag: maths

Questions Related to maths

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. $0< a< 100$

Show answer
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. $\dfrac {11}2$

Show answer
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. $1$

Show answer
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
$\dfrac{{\left(1-r\right)}^{2}}{1-{r}^{2}}=\dfrac{400}{100}=4$
$\Rightarrow \dfrac{{\left(1-r\right)}^{2}}{\left(1-r\right)\left(1+r\right)}=4$
$\Rightarrow \dfrac{1-r}{1+r}=4$
$\Rightarrow 4-4r=1+r$
$\Rightarrow 5r=3$
$\therefore r=\dfrac{3}{5}$

For first $n$ natural numbers we have the following results with usual notations $ \displaystyle \sum _{r=1}^{n}r =\frac{n(n+1)}{2}, \sum _{r=1}^{n}r^{2} =\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^{n}r^{3}=\left ( \sum _{r=1}^{n}r \right )^{2}$ If $\displaystyle a _{1}a _{2}....a _{n} \in A.P $ then sum to $n$ terms of the sequence $\displaystyle \frac{1}{a _{1}a _{2}},\frac{1}{a _{2}a _{3}},...\frac{1}{a _{n-1}a _{n}}$ is equal to $\displaystyle \frac{n-1}{a _{1}a _{n}}$
 and the sum to $ n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by  $\displaystyle S _{n}= \frac{lr-a}{r-1}$ for $ r \neq 1 $ for $ r =1 $ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
 $\displaystyle \frac{lr-a}{r-1} $ when $\displaystyle n \rightarrow \infty ,\left |  r \right | < l $ where $l$ is the last term of $G.P.$  On the basis of above data answer the following questionsThe sum to infinite terms of the series $\displaystyle \frac{1}{2}+\frac{1}{6}+\frac{1}{18}+.. $ is equal to ?

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

  1. $\displaystyle \frac{4}{3}$

  2. $\displaystyle \frac{3}{4}$

  3. $\displaystyle \frac{8}{3}$

  4. Does not exit

Show answer
Correct Option: B
Explanation:

Let, ${ S } _{ \infty  }=\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 6 } +d\frac { 1 }{ 18 } +..\infty $

$\Rightarrow { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( 1+\dfrac { 1 }{ 3 } +\dfrac { 1 }{ { 3 }^{ 2 } } +....\infty  \right) $

As we know that, sum of infinite G.P series $=\dfrac { a }{ 1-r } $

Therefore, $ { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ 1-\left( 1/3 \right)  }  \right) =\dfrac { 3 }{ 4 } $

Ans: B

If $\displaystyle x=\sum _{a=0}^{\infty }a^{n},y=\sum _{a=0}^{\infty }b^{n},z=\sum _{a=0}^{\infty }c^{n}$ Where $a,b,c $ are in A.P and $\displaystyle \left | a \right |<1,\left | b \right |<1,\left | c \right |<1$ then $x,y,z$ are in

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

  1. H.P

  2. Arithmetic-Geometric progression

  3. A.P

  4. G.P

Show answer
Correct Option: A
Explanation:

Given $\displaystyle \left | a \right |< 1,\left | b \right |< 1,\left | c \right |< 1,\ \ a,b,c \in A.P$


and $\displaystyle \sum _{n=0}^{\infty }a^{n}=\frac{1}{1-a},\sum _{n=0}^{\infty }b^{n}=\frac{1}{1-b},\sum _{r=0}^{\infty }c^{n}=\frac{1}{1-c}$

$\displaystyle \therefore x=\frac{1}{1-a},y=\frac{1}{1-b},c=\frac{1}{1-c}$

$\displaystyle \Rightarrow a=\frac{x-1}{x},b=\frac{y-1}{y},c=\frac{z-1}{z}$

$\displaystyle \because 2b=a+c \ as \ a,b,c \in A.P$

$\displaystyle 2\left ( \frac{y-1}{y} \right )=\frac{x-1}{x}+\frac{z-1}{z}\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$\displaystyle \Rightarrow x,y,z \in H.P$

If $R \subset\left ( 0,\pi  \right )$ denote the set of values of which satisfies the equation $ \displaystyle 2^{\left ( 1+\left | \cos x \right |+\left | cos^{2}x \right |+\left | cos^{3}x \right | \right )+\left | cos^{4}x  \right |...............\infty}=4$ then $R$ equals

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

  1. $\displaystyle\left { -\frac{\pi }{3} \right }$

  2. $\displaystyle\left { \frac{\pi }{3},\frac{2\pi }{3} \right }$

  3. $\displaystyle\left { \frac{-\pi }{3},\frac{2\pi }{3} \right }$

  4. $\displaystyle\left { \frac{\pi }{3},\frac{-2\pi }{3} \right }$

Show answer
Correct Option: B
Explanation:

${ 2 }^{ \left( 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty  \right)  }={ 2 }^{ 2 }\ \Rightarrow 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty =2\ \Rightarrow \dfrac { 1 }{ 1-\left| \cos { x }  \right|  } =2\ \Rightarrow 1-\left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow \left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow x=\dfrac { \pi  }{ 3 } ,\dfrac { 2\pi  }{ 3 } $
  in the range $\left( 0,\pi  \right) $

The sum of the series
$\dfrac { 1 } { 1.2 } - \dfrac { 1 } { 2.3 } + \dfrac { 1 } { 3.4 } \ldots \ldots \ldots$  up to  $\infty$  is equal to

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

  1. $\log _{ { { e } } } \left( \dfrac { 4 }{ { e } } \right) $

  2. $2 \log _ { e } 2$

  3. $\log _ { e } 2 - 1$

  4. $\log _ { e } 2$

Show answer
Correct Option: A

The sum of the infinite series, ${ 1 }^{ 2 }-\frac { { 2 }^{ 2 } }{ 5 } +\frac { { 3 }^{ 2 } }{ { 5 }^{ 2 } } -\frac { { 4 }^{ 2 } }{ { 5 }^{ 3 } } +\frac { { 5 }^{ 2 } }{ { 5 }^{ 4 } } -\frac { { 6 }^{ 2 } }{ { 5 }^{ 5 } } +.........$ is :

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

  1. $\frac { 1 }{ 2 } $

  2. $\frac { 25 }{ 24 } $

  3. $\frac { 25 }{ 54 } $

  4. $\frac { 125 }{ 252 } $

Show answer
Correct Option: A

The first term of an infinitely decreasing G.P. is unity and its sum 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. $\displaystyle \frac {S}{2S-1}$

  2. $\displaystyle \frac {S^2}{2S-1}$

  3. $\displaystyle \frac {S}{2-S}$

  4. $S^2$

Show answer
Correct Option: B
Explanation:

Let common ratio is $r<1$
Then G.P is $1,r,{ r }^{ 2 },{ r }^{ 3 },...\infty $
$S=1+r+{ r }^{ 2 }+{ r }^{ 3 }+...\infty $
$\displaystyle \Rightarrow S=\frac { 1 }{ 1-r } $
Then G.P formed by squaring the terms 
$1,{ r }^{ 2 },{ r }^{ 4 },{ r }^{ 6 },...\infty $
$\displaystyle { S }'=\frac { 1 }{ 1-{ r }^{ 2 } } =\frac { 1 }{ \left( 1-r \right) \left( 1+r \right)  } =\frac { { S }^{ 2 } }{ 2S-1. } $