Tag: maths

Questions Related to maths

 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, $ac+bc=ab+c$

Sum to infinity of the series $\displaystyle \frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ is

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

  1. $\displaystyle \frac { 4 }{ 9 } $

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

  3. $\displaystyle \frac { 2 }{ 9 } $

  4. none of these

Show answer
Correct Option: C
Explanation:

Let $\displaystyle S=\frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ to $\infty$   ...(1)


Multiplying both sides by $\displaystyle -\frac { 1 }{ 2 } $, the common ratio $G.P.$


$\displaystyle -\frac { 1 }{ 2 } S=-\frac { 2 }{ 6 } +\frac { 5 }{ 12 } -\frac { 8 }{ 24 } +...$ to $\infty$    ....(2)

Subtracting (2) from (1), we have

$\displaystyle \frac { 3 }{ 2 } S=\frac { 2 }{ 3 } -\frac { 3 }{ 6 } +\frac { 3 }{ 12 } -\frac { 3 }{ 24 } +...$ to $\infty$

$\displaystyle =\frac { 2 }{ 3 } -\left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +... \right) $

$\displaystyle =\dfrac { 2 }{ 3 } -\dfrac { \dfrac { 1 }{ 2 }  }{ 1-\left( -\dfrac { 1 }{ 2 }  \right)  } =\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 } $

$\displaystyle \therefore S=\frac { 1 }{ 3 } \times \frac { 2 }{ 3 } =\frac { 2 }{ 9 } $

If $S$ is the sum to infinity of a GP, whose first term is $a$, then the sum of the first $ n$  terms is

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

  1. $\displaystyle S\left ( 1-\frac{a}{S} \right )^{n}$

  2. $\displaystyle S\left [ 1-\left ( 1-\frac{a}{S} \right )^{n} \right ]$

  3. $\displaystyle a\left [ 1-\left ( 1-\frac{a}{S} \right )^{n} \right ]$

  4. none of these

Show answer
Correct Option: B
Explanation:

Let r be the common ration.
GIven, $S _\infty=S=\frac{a}{1-r}$
$\Rightarrow 1-r=\frac aS$
$\Rightarrow r=1-\frac aS$
Now, sum of n terms is given by
$S _n=\dfrac{a(1-r^n)}{1-r}$


       $=\dfrac{a(1-(1-\frac aS)^n)}{1-(1-\frac aS)}$

       $=\dfrac{a(1-(1-\frac aS)^n)}{\frac aS}$


       $=S[1-(1-\frac aS)^n]$
Option B is correct.

$\displaystyle2+1+\frac{1}{2}+\frac{1}{4}+\cdots\cdots\infty$ is

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

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: D
Explanation:

The given series is a Geometric Progression, with first terms $ a =2$ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {1}{2} $

For a GP, sum to infinity is given by the formula $ \dfrac {a}{1-r} $

So, for the given series, $ S _\infty  = \dfrac {2}{1-\dfrac {1}{2}} = 4 $

What is the sum of the infinite geometric series where the beginning term is $2$ and the common ratio is $3$?

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

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$

Show answer
Correct Option: B
Explanation:

From the given information, we have
first term $=a=2 $, common ratio $r=3$
We know $S = \dfrac{a}{1-r}$
Therefore, $S = \dfrac{2}{1-3}$
$\Rightarrow S = -1$

The value of the infinite product $6^{\frac{1}{2}}\times 6^{\frac{1}{2}}\times 6^{\frac{3}{8}}\times 6^{\frac{1}{4}}\times .........$ is

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

  1. 6

  2. 36

  3. 216

  4. $\infty$

Show answer
Correct Option: B
Explanation:
$6^{\frac12}\times6^{\frac12}\times6^{\frac38}\times6^{\frac14}\times...............$
$=6^{\frac12+\frac24+\frac38+\frac4{16}+...................}$
$=6^{\frac12\left(1+\frac22+\frac3{2^2}+\frac4{2^3}+......................\right)}$

Let $S=1+\cfrac22+\cfrac3{2^2}+\cfrac4{2^3}+............$, then
$\cfrac S2=S-\cfrac S2$
or, $\cfrac S2=1+\cfrac12+\cfrac1{2^2}+\cfrac1{2^3}+...........$
or, $\cfrac S2=\cfrac1{1-\cfrac12}$ ..... [Using sum of infinite terms of an G.P]
or, $S=4$
Then we have,
$6^{\frac12}\times6^{\frac12}\times6^{\frac38}\times6^{\frac14}\times...............$
$=6^{\frac12\times4}=6^2=36$
Hence, B is the correct option.

Calculate the sum of the infinite series: $1 - \dfrac {1}{3} + \dfrac {1}{9} - \dfrac {1}{27} + .....$.

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

  1. $\dfrac {2}{3}$

  2. $\dfrac {3}{4}$

  3. $1$

  4. $\dfrac {4}{3}$

  5. $\dfrac {3}{2}$

Show answer
Correct Option: B
Explanation:

Given series is $1,-\dfrac{1}{3},+\dfrac{1}{9},-\dfrac{1}{27}......................$

Then common ratio $=$ $ (-\dfrac{1}{3})/1=-\dfrac{1}{3}$
Then sum of infinite series $S=$ $\dfrac{a _{1}}{1-r}=\dfrac{1}{1-(-\frac{1}{3})}=\dfrac{1}{\frac{1+3}{3}}=\dfrac{3}{4}$