Tag: sum to infinite terms of a gp

Questions Related to sum to infinite terms of a gp

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-

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

  1. $\mid x\mid > 5$

  2. -5 < x < 5

  3. $\mid x\mid < 5/2$

  4. $\mid x\mid > 5/2$

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Correct Option: A
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 $\mid 5/x\mid < 1$
$\Leftrightarrow \mid x\mid > 5$

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

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

  1. True

  2. False

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Correct Option: A

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

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

  1. $0$

  2. $1$

  3. $x$

  4. $\infty$

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Correct Option: C
Explanation:

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


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

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

$=x^{1}$ 

$=x$

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

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

  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

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Correct Option: C
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.

The sum of $7+1+.......$

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

  1. $\dfrac{49}{6}$

  2. $\dfrac{49}{8}$

  3. $\dfrac{49}{14}$

  4. None of these

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Correct Option: A
Explanation:

$7+1+\dfrac 17 ...$

Forms GP  with $a=7\r=\dfrac 17 $
So $S _{\infty}=\dfrac{7}{1-\dfrac 17}\\dfrac{49}{6}$

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

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

  1. $x<3$

  2. $x>3$

  3. $x=0$

  4. $x=-3$

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Correct Option: A
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
$r < 1$
$\therefore \dfrac{2x}{x+3} < 1$
$\therefore 2x < x+3$
$\therefore x < 3$

Find the sum of $4,2,1,\cdots$ 

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

  1. 8

  2. 16

  3. 32

  4. 64

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Correct Option: A
Explanation:

The series $4,2,1,\cdots$ form GP

With $a=4$ and $r=\dfrac 12$
$S _{\infty}=\dfrac{a}{1-r}\\dfrac{4}{1-\dfrac 12 }=\dfrac{4}{\dfrac 12}=4\times 2=8$

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

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

  1. $x^{1/3}$

  2. $x^{2/3}$

  3. $x^{1/2}$

  4. $x$

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Correct Option: C
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.

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

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

  1. $\dfrac{7}{16}$

  2. $\dfrac{9}{16}$

  3. $\dfrac{1}{9}$

  4. $\dfrac{7}{9}$

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Correct Option: A
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, $\dfrac{\dfrac{3}{4}}{1-r}=\dfrac{4}{3}\Rightarrow\,r=1-\dfrac{9}{16}=\dfrac{7}{16}$

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

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

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

  2. $2cos144^{ \circ }$

  3. $2sin18^{ \circ }$

  4. none

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Correct Option: A
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 
$x=\dfrac{-1\pm\sqrt{1+4}}2\2\dfrac{-1\pm\sqrt5}{4}\2\cos 36^{\circ}$