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Questions Related to maths

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

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

  1. $0< a< 130$
  2. $0< a< 30$
  3. $0< a< 15$
  4. $0< a< 100$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $
$\Rightarrow -1 < -r < 0.$
$\Rightarrow  0 < 1-r < 1$
$\therefore  0 < 15(1-r) < 15$
$\Rightarrow 0 < a < 15$

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

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

  1. $x^2-8x+16=0$
  2. $x^2-3x+2=0$
  3. $x^2-5x+4=0$
  4. $x^2+5x+4=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

The sum of  $3,1,\dfrac 13 ,....$ is

  1. $\dfrac 52$
  2. $\dfrac 92$
  3. $\dfrac 72$
  4. $\dfrac {11}2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $\dfrac{a}{1-r}$

$\implies  \dfrac 3{1-\dfrac 13}$

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


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

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

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{3}{5}$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$\dfrac{{\left(1-r\right)}^{2}}{1-{r}^{2}}=\dfrac{400}{100}=4$
$\Rightarrow \dfrac{{\left(1-r\right)}^{2}}{\left(1-r\right)\left(1+r\right)}=4$
$\Rightarrow \dfrac{1-r}{1+r}=4$
$\Rightarrow 4-4r=1+r$
$\Rightarrow 5r=3$
$\therefore r=\dfrac{3}{5}$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

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 ?

  1. $\displaystyle \frac{4}{3}$
  2. $\displaystyle \frac{3}{4}$
  3. $\displaystyle \frac{8}{3}$

  4. Does not exit
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. H.P

  2. Arithmetic-Geometric progression

  3. A.P

  4. G.P
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

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

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

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

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

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

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

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

  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. $\displaystyle\left \{ \frac{\pi }{3},\frac{-2\pi }{3} \right \}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

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

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

  1. $\log _{ { { e } } } \left( \dfrac { 4 }{ { e } } \right) $
  2. $2 \log _ { e } 2$
  3. $\log _ { e } 2 - 1$
  4. $\log _ { e } 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

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 :

  1. $\frac { 1 }{ 2 } $
  2. $\frac { 25 }{ 24 } $
  3. $\frac { 25 }{ 54 } $
  4. $\frac { 125 }{ 252 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is an arithmetico-geometric series of the form sum(n^2 * r^(n-1)). The sum can be found using the method of differences or differentiation of geometric series. The result for this specific series is 1/2.

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

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

  1. $\displaystyle \frac {S}{2S-1}$
  2. $\displaystyle \frac {S^2}{2S-1}$
  3. $\displaystyle \frac {S}{2-S}$
  4. $S^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. } $