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

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
$r < 1$
$\therefore \dfrac{2x}{x+3} < 1$
$\therefore 2x < x+3$
$\therefore x < 3$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $, then $y =$

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

$y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $
 $ =x^{\cfrac{1}{3}+\cfrac{1}{3^2}+\cfrac{1}{3^3}+........\infty }=x^{\cfrac{1/3}{1-1/3}}=x^{1/2}$
Hence, option 'C' is correct.

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

If sum of an infinite geometric series is $\dfrac{4}{3}$ and its Ist term is $\dfrac{3}{4}$, then its common ratio is

  1. $\dfrac{7}{16}$
  2. $\dfrac{9}{16}$
  3. $\dfrac{1}{9}$
  4. $\dfrac{7}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that sum of infinite geometric series $=\dfrac{a}{1-r}$
Where $a=\text{first term}$ and $r=\text{common ratio}$.
$\dfrac{a}{1-r}=\dfrac{4}{3}$
Then, $\dfrac{\dfrac{3}{4}}{1-r}=\dfrac{4}{3}\Rightarrow\,r=1-\dfrac{9}{16}=\dfrac{7}{16}$

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

The value of x that satisfies the relation 
$x=1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }-{ x }^{ 5 }+........\infty $ 

  1. $2cos{ 3 }6^{ \circ }$
  2. $2cos144^{ \circ }$
  3. $2sin18^{ \circ }$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series $1-x+x^2-....$ form $GP$ with $a=1 ,r=-x$

Sum of infinte GP is $x=\dfrac{a}{1-r}\x=\dfrac{1}{1+x}\x+x^2=1\x^2+x-1=0$
By quadratic formulae 
$x=\dfrac{-1\pm\sqrt{1+4}}2\2\dfrac{-1\pm\sqrt5}{4}\2\cos 36^{\circ}$