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Questions Related to maths

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).

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

If the sum of the series $2+\frac {\displaystyle 5}{\displaystyle x}+\frac {\displaystyle 25}{\displaystyle x^2}+\frac {\displaystyle 125}{\displaystyle x^3}+....$ is finite, then-

  1. $\mid x\mid > 5$

  2. -5 < x < 5
  3. $\mid x\mid < 5/2$
  4. $\mid x\mid > 5/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can rewrite the series as
$1+1+\dfrac {5}{x}+(\dfrac {5}{x})^2+(\dfrac {5}{x})^3+.....$
We can sum up this series if $\mid 5/x\mid < 1$
$\Leftrightarrow \mid x\mid > 5$

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

If $x=1+a+a^2+...\infty$ where $|a| <1 $ and $y=1+b+b^2+...\infty$, where $|b| < 1$, then $1+ab+a^2b^2+...\infty =\dfrac{xy}{x+y-1}$.

  1. True

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

This is the same identity derived in question 466317. The sum of the series 1 + ab + a^2b^2 + ... is indeed xy / (x + y - 1) given the definitions of x and y.

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

${x}^{\cfrac{1}{2}}.{x}^{\cfrac{1}{4}}.{x}^{\cfrac{1}{8}}.{x}^{\cfrac{1}{16}}.....$ to $\infty$

  1. $0$
  2. $1$
  3. $x$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^{1/2}.x^{1/4}......\infty =x^{\left ( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} \right )}$ 


$=x\left ( 1/2+1/2^{2}+1/2^{3}+1/2^{4}+.......\infty  \right )$

$=x^{\frac{1}{2}\left ( \frac{1-(1/2)\infty }{1-1/2} \right )}$ 

$=x^{1}$ 

$=x$

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

The solution of the equation $(8)^{1+|cos x|+|cos x|^2+|cos x|^3+...)}=4^3$ in the interval $(-\pi, \pi)$ are.

  1. $\pm \dfrac {\pi }{3}, \pm \dfrac {\pi }{6}$
  2. $\pm \dfrac {\pi }{3}, \pm {\pi }$
  3. $\pm \dfrac {\pi }{3}, \pm \dfrac {2\pi }{3}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 $(8)^{1+|cos x|+|cos x|^2+|cos x|^3+...)}=4^3$
$\Rightarrow (8)^{\dfrac{1}{1-|\cos x|}}=4^3=64=8^2$, since $|\cos x| < 1$ in $(-\pi, \pi)$
$\Rightarrow {\dfrac{1}{1-|\cos x|}}=2$
$\Rightarrow |\cos x|=\cfrac{1}{2}$
The solution in the given interval is,
$x=\pm \cfrac{\pi}{3}, \pm \cfrac{2\pi}{3}$
Hence, option 'C' is correct.

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

The series $\dfrac{2x}{x+3}+(\dfrac{2x}{x+3})^{2}+(\dfrac{2x}{x+3})^{3}+........\infty$ will have a definite sum when  

  1. $x<3$
  2. $x>3$
  3. $x=0$
  4. $x=-3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{2x}{x+3}+\left(\dfrac{2x}{x+3}\right)^{2}+\left(\dfrac{2x}{x+3}\right)^{3}.......\infty $
$\therefore a=\dfrac{2x}{x+3}$   $r=\dfrac{2x}{x+3}$
$\therefore s=\dfrac{a}{1-r}=\dfrac{\dfrac{2x}{x+3}}{1-\dfrac{2x}{x+3}}$
$=\dfrac{2x}{x+3-2x}=\dfrac{2x}{3-x}$
Now, to have definite sum
$r < 1$
$\therefore \dfrac{2x}{x+3} < 1$
$\therefore 2x < x+3$
$\therefore x < 3$