Tag: sum to infinite terms of a gp

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Questions Related to sum to infinite terms of a gp

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

The value of $3 - 1 + \frac{1}{3} - \frac{1}{9} +  \ldots $ is equal to

  1. $\dfrac{{20}}{9}$
  2. $\dfrac{{9}}{20}$
  3. $\dfrac{{9}}{4}$
  4. $\dfrac{{4}}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The series is 3 - 1 + 1/3 - 1/9 + ... which can be split into 3 + (-1 + 1/3 - 1/9 + ...). The part in parentheses is a geometric series with first term a = -1 and common ratio r = -1/3. The sum is a / (1 - r) = -1 / (1 - (-1/3)) = -1 / (4/3) = -3/4. Adding the initial 3 gives 3 - 3/4 = 9/4.

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

Let $P = 3^{1/3} . 3^{2/9} . 3^{3/27} ...\infty$, then $P^{1/3}$ is equal to

  1. $3^{2/3}$
  2. $\sqrt {3}$
  3. $3^{1/3}$
  4. $3^{1/4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Above in an infinite A.G.S. with $a = 1, d = 1$ for A.P., $b = \dfrac {1}{3}, r = \dfrac {1}{3}$ for G.P.
$\therefore S _{\infty} = \dfrac {ab}{1 - r} + \dfrac {dbr}{(1 - r)^{2}} = \dfrac {\dfrac {1}{3}}{1 - \dfrac {1}{3}} + \dfrac {1 . \dfrac {1}{3} . \dfrac {1}{3}}{\left (1 - \dfrac {1}{3}\right )^{2}} = \dfrac {1}{2} + \dfrac {1}{4} = \dfrac {3}{4}$
$\therefore P = 3^{S} = 3^{3/4} \therefore P^{1/3} = 3^{1/4}$.

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

The value of $9^\cfrac{1}{3}.9^\cfrac{1}{9}.9^\cfrac{1}{27}...........$ upto $\infty$, is

  1. $1$
  2. $3$
  3. $9$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${9^{\cfrac{1}{3}}}{.9^{\cfrac{1}{9}}}{.9^{\cfrac{1}{{27}}}} -  -  -  - upto\,\,\infty $
$ = {9^{\left( {\cfrac{1}{3} + \cfrac{1}{9} + \cfrac{1}{{27}} +  -  -  - } \right)}}$
$ = {9^{\left( {\cfrac{{\cfrac{1}{3}}}{{1 - \cfrac{1}{3}}}} \right)}}$
$ = {9^{\left( {\cfrac{{\cfrac{1}{3}}}{{\cfrac{2}{3}}}} \right)}}$
$ = {9^{\cfrac{1}{2}}}$
$ = 3$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x=1+a+{ a }^{ 2 }+{ a }^{ 3 }+....$ to $\infty \left( \left| a \right| <1 \right) $ and 
$y=1+b+{ b }^{ 2 }+{ b }^{ 3 }+...$ to $\infty \left( \left| b \right| <1 \right) $ then
$1+ab+{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 3 }{ b }^{ 3 }+...$ to $\infty =\cfrac { xy }{ x+y-1 } $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ x }=\dfrac { 1 }{ 1-a } $ ${ y }=\dfrac { 1 }{ 1-b } $ [summing infinite $G.P's$].
$\therefore a=\dfrac { x-1 }{ x } $, $b=\dfrac { y-1 }{ y } $
$\therefore 1+ab+{ a }^{ 2 }{ b }^{ 2 }+...\infty $
$=\dfrac { 1 }{ 1-ab } =\dfrac { 1 }{ 1-\dfrac { (x-1)(y-1) }{ xy }  } =\dfrac { xy }{ x+y-1 }. $

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

The sum to infinity of the series $1 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ......,is$

  1. $3$
  2. $4$
  3. $6$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $S$=$1 + \cfrac{2}{3} + \cfrac{6}{{{3^2}}} + \cfrac{{10}}{{{3^3}}} + ......$

$\cfrac{S}{3} = \cfrac{1}{3} + \cfrac{2}{{{3^2}}} + \cfrac{6}{{{3^3}}} + ....$

$S - \cfrac{S}{3} = 1 + \cfrac{1}{3} + \cfrac{4}{{{3^2}}} + ....$

$\cfrac{{2S}}{3} = \cfrac{{\cfrac{4}{3} }}{{1 - \cfrac{1}{3}}}$

$ = \dfrac{\cfrac{4}{3}} { \cfrac{2}{3}} = \cfrac{4}{2} = 2$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x = 1\, + a + {a^2} + ......\infty $, $y = 1\, + b + {b^2}\,\, + ......\infty $ where $\left| a \right| < 1$ and $\left| b \right| < 1$, then $\left( {1 + ab + {a^2}{b^2} + ........\infty } \right) = ?$

  1. $\frac{xy}{x+y}$
  2. $\frac{x+y}{xy}$
  3. $\frac{xy}{x+y+1}$
  4. $\frac{xy}{x+y-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given x = 1/(1-a) and y = 1/(1-b), we have a = (x-1)/x and b = (y-1)/y. The series 1 + ab + a^2b^2 + ... is a geometric series with sum 1/(1-ab). Substituting a and b: 1 / (1 - ((x-1)/x)((y-1)/y)) = 1 / (1 - (xy - x - y + 1)/(xy)) = xy / (xy - xy + x + y - 1) = xy / (x + y - 1).

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

Value of $y = {\left( {0.64} \right)^{{{\log } _{0.25}}\left( {\cfrac{1}{3} + \cfrac{1}{{{3^2}}} + \cfrac{1}{{{3^3}}}....upto   \infty } \right)}}$ is :

  1. $0.9$
  2. $0.8$
  3. $0.6$
  4. $0.25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$y= (0.64)^{log _{0.25} \left(\dfrac{1}{3}+ \dfrac{1}{3^{2}}+ \dfrac{1}{3^{3}}+..... \right)}$
$=(0.64)^{\log _{0.25}^{\left( \dfrac{\dfrac{1}{3}}{1-1/3} \right)}}$
$=(0.64)^{\log _{0.25} } \left( \dfrac{1}{2} \right)$
$= (0.64)^{\log 0.5} _{0.25}$
$(0.64)^{0.5}= (0.64)^{1/2}= \sqrt{0.64}= 0.8$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $y=x-x^2+x^3-x^4+....\infty$, then value of x will be?

  1. $y+\dfrac{1}{y}$
  2. $\dfrac{y}{1+y}$
  3. $y-\dfrac{1}{y}$
  4. $\dfrac{y}{1-y}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series y = x - x^2 + x^3 - x^4 + ... is a geometric series with first term a = x and common ratio r = -x. The sum is y = x / (1 - (-x)) = x / (1 + x). Solving for x: y(1 + x) = x, so y + xy = x, which means y = x - xy = x(1 - y). Thus, x = y / (1 - y).