Tag: sum to infinite terms of a gp

Questions Related to sum to infinite terms of a gp

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. } $

 If  $0<\phi < \pi /2,$   and
 $x= \sum _{n=0}^{\infty} \cos ^{2n} \phi$, $ y=\sum _{n=0}^{\infty } \sin ^{2n} \phi$                     
and $z=\sum _{n=0}^{\infty} \cos ^{2n} \phi \sin ^{2n} \phi $ 
then

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

  1. xyz $=$xz+y

  2. xyz$=$xy+z

  3. xyz$=$x+y+z

  4. xy$=$yz+z

Show answer
Correct Option: B,C
Explanation:

The series x converges to a value $\dfrac{1}{1-{\cos^2 \phi}}\;=\;{cosec^2 \phi}$

Similarly y converges to  a value ${sec^2 \phi} $

In a similar fashion.. z converges to $\dfrac{{cosec^2 \phi}{sec^2 \phi}}{{cosec^2 \phi}{sec^2 \phi} - 1}$
 $ z = \dfrac{xy}{xy - 1} $

$xy + z = xyz&gt;$; Option B.
And, also
$x + y = xy$
$\therefore xyz = x+ y+ z&gt;$; Option C.

Find the sum of the infinite geometric series where the beginning term is $-1$ and the common ratio is $\dfrac{1}{2}$.

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

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$

Show answer
Correct Option: D
Explanation:

Given that first term $a=-1$ and common ratio is $r=\dfrac{1}{2}$

We know $\text{sum} = \dfrac{a}{1-r}$
$\Rightarrow \text{sum} = \dfrac{-1}{1-\frac{1}{2}}$
$\Rightarrow \text{sum} = \dfrac{-1}{\frac{1}{2}}$
$\Rightarrow \text{sum} = -2$

$1 + x + x^2 + x^3 +......$ = ?

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

  1. $\dfrac{1}{1-x}$

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

  3. $\dfrac{1}{1-x^3}$

  4. $\dfrac{x}{1-x}$

Show answer
Correct Option: A
Explanation:

Sum of a GP $a,ar,ar^2......$

$=\dfrac{a(r^n -1)}{r-1}$    (for $n$ terms)
For the given series,
$a=1, r=x, n \to \infty$
Sum of the series is:
$\text{sum} = \lim _{n \to \infty} \dfrac{x^n-1}{x-1}$
For $x>1$ 
$\text{sum} \to \infty$
For $x<1$
$x^n \to 0$
$\Rightarrow \text{sum} = \dfrac{-1}{x-1}$
$\Rightarrow \text{sum} = \dfrac{1}{1-x}$

If $a=\sum _{ n=0 } ^{\infty  }{x^n } ,b=\sum _{n=0  }^{ \infty  }{ y^n } , c=\sum _{n=0  }^{ \infty  }{ (xy)^n } $ where $|x| ,| y| < 1$ ; then

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

  1. $abc = a + b + c$

  2. $ab + bc = ac + b$

  3. $ac + bc = ab + c$

  4. $ab + ac = bc + a$

Show answer
Correct Option: C
Explanation:

Clearly every summation is infinite series,
$a=\cfrac{1}{1-x}, b = \cfrac{1}{1-y}$ and $c=\cfrac{1}{1-xy}$
or $x=1-\cfrac{1}{a}, y=1-\cfrac{1}{b}$
Simplifying above equation we get, $ac+bc=ab+c$