Tag: sequence, progression and series

Questions Related to sequence, progression and series

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


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

  1. H.P

  2. Arithmetic-Geometric progression

  3. A.P

  4. G.P


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

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


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

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

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

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

  4. $\log _ { e } 2$


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 :

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

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

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

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


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

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

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

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

  4. $S^2$


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

  1. xyz $=$xz+y

  2. xyz$=$xy+z

  3. xyz$=$x+y+z

  4. xy$=$yz+z


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 $\dfrac{1}{2}$.

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$


Correct Option: D
Explanation:

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

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

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

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

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

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

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


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

  1. $abc = a + b + c$

  2. $ab + bc = ac + b$

  3. $ac + bc = ab + c$

  4. $ab + ac = bc + a$


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, $ac+bc=ab+c$