Tag: understanding geometric progressions

Questions Related to understanding geometric progressions

If $\alpha, \beta, \gamma$ are non-constant terms in G.P and equations $\alpha { x }^{ 2 }+2\beta x+\gamma =0\quad $ and ${x}^{2}+x-1=0$ has a common root then $\left( \gamma -\alpha  \right) ,\beta $ is

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $\alpha \beta $

  2. $\beta \gamma $

  3. $\gamma \alpha $

  4. $0$

Show answer
Correct Option: C
Explanation:

Let the common ratio of G.P is $r$ Therefore $\quad \beta =\alpha t,\alpha { t }^{ 2 }$
Equation $\alpha { x }^{ 2 }+2\alpha rx+\alpha { t }=0\quad 
\Rightarrow { x }^{ 2 }+2rx+{ t }^{ 2 }=0....(i)$
Given equation (i) and ${ x }^{ 2 }+x-1=0....(ii)$ has a common root
$(i)-(ii)\Rightarrow (2e-1)x+({ r }^{ 2 }+1)=0\Rightarrow x=\cfrac { -\left( { r }^{ 2 }+1 \right)  }{ 2r-1 } ....(iii)\quad $
Putting (iii) in equation (ii) $\Rightarrow { \left( { r }^{ 2 }+1 \right)  }^{ 2 }-\left( { r }^{ 2 }+1 \right) (2r-1)-{ \left( { 2r }^{ 2 }-1 \right)  }^{ 2 }=0\Rightarrow { r }^{ 4 }-2{ r }^{ 3 }-{ r }^{ 2 }+2r+1=0....(iv)$
dividing equation (iv) by ${r}^{2}$ $\Rightarrow { \left( r-\cfrac { 1 }{ r }  \right)  }^{ 2 }-2{ \left( r-\cfrac { 1 }{ r }  \right)  }+1=0\Rightarrow { \left( r-\cfrac { 1 }{ r } -1 \right)  }^{ 2 }=0\Rightarrow \cfrac { r-1 }{ r } =1....(v)\quad $
$\left( \gamma -\alpha  \right) \beta =\left( \alpha { r }^{ 2 }-\alpha  \right) \times \alpha r={ \alpha  }^{ 2 }\left( { \alpha  }^{ 2 }-1 \right) r={ \alpha  }^{ 2 }(r-1)={ \alpha  }^{ 2 }{ r }^{ 2 }$
(using $(v)=\alpha \times \alpha { t }^{ 2 }\quad $

Write down the first five terms of the geometric progression which has first term 1 and common ratio 4.

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. 1, 4, 16, 64, 244

  2. 1, 4, 24, 64, 256

  3. 1, 4, 16, 32, 256

  4. 1, 4, 16, 64, 256

Show answer
Correct Option: D
Explanation:

Let a and d be the first term and common ratio of the GP respectively.
Given a=1 and d=4.
Now, $a _n=ar^{n-1}$
$\therefore a _1=a=1$
$a _2=ar=1\times4=4$
$a _3=ar^2=1\times(4)^2=16$
$a _4=ar^3=1\times(4)^3=64$
$a _5=ar^4=1\times(4)^4=256$





$\displaystyle \frac{1}{c},(\frac{1}{ca})^{\dfrac{1}{2}},\frac{1}{a}$ is in

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. AP

  2. GP

  3. HP

  4. NONE

Show answer
Correct Option: B
Explanation:

Given series


$\dfrac{1}{c},\left(\dfrac{1}{ca}\right)^{\dfrac{1}{2}},\dfrac{1}{a}$

Lets consider a G.P of elements $A,B,C$

 $\therefore$ Geo.mean $\Rightarrow B^2=AC$

Comparing it with given series.

$A=\dfrac{1}{c}B=\left(\dfrac{1}{ca}\right)^{\dfrac{1}{2}},C=\dfrac{1}{a}$

$\therefore B^2=\left(\dfrac{1}{ca}\right)^{\dfrac{1}{2}\times 2}$

            $=\dfrac{1}{ca}$........(1)

$AC=\dfrac{1}{c}\times \dfrac{1}{a}=\dfrac{1}{ca}$..............(ii)

$\therefore (i)=(ii)$

$\therefore B^2=AC$ So given series is in G.P 

Determine the relations among x, y and z if $y^{2}=xz$

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. A.P

  2. G.P

  3. A.G.P

  4. none of these

Show answer
Correct Option: B
Explanation:

It is fundamental concept that if $y^2 = xz$, then $x,y,z\in$  G.P

Find the sum the infinite G.P.: $\displaystyle {\frac{2}{3}\, -\, \frac{4}{9}\, +\, \frac{8}{27}\, -\, \frac{16}{21}\, +\, ........}$ 

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $\displaystyle \frac{2}{5}$

  2. $\displaystyle \frac{3}{5}$

  3. $\displaystyle \frac{19}{27}$

  4. $\displaystyle \frac{8}{5}$

Show answer
Correct Option: A
Explanation:

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=\dfrac{2}{3} ,r=-\dfrac{2}{3}$
$\therefore S _{\infty }=\dfrac{\frac{2}{3}}{1-\left ( -\frac{2}{3} \right )}$


$\Rightarrow \dfrac{\frac{2}{3}}{1+\frac{2}{3}}$

$\Rightarrow \dfrac{\frac{2}{3}}{\frac{5}{3}}$

$\Rightarrow \dfrac{2}{5}$

Sum the series: $\displaystyle {1\, -\, \frac{1}{3}\, +\, \frac{1}{3^2}\, -\, \frac{1}{3^3}\, +\, \frac{1}{3^4}.......\infty}$

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $\displaystyle \frac{3}{4}$

  2. $\displaystyle \frac{4}{3}$

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

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

Show answer
Correct Option: A
Explanation:

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=1 ,r=-\dfrac{1}{3}$
$\therefore S _{\infty }=\dfrac{1}{1-\left ( -\dfrac{1}{3} \right )}$
$= \dfrac{1}{1+\dfrac{1}{3}}$


$ =\dfrac{1}{\dfrac{4}{3}}$

$= \dfrac{3}{4}$

If a, b and c are in geometric progression, then $a^2$, $b^2$ and $c^2$ are in _____ progression.

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. AP

  2. GP

  3. HP

  4. AGP

Show answer
Correct Option: B
Explanation:

Given $ a,b,c $ are in GP.
So, the common ratio between the first and second term ; second and third will be the same.
$ => \dfrac {b}{a} = \dfrac {c}{b} $


$ => b^2 = ac $

If we square both sides, we see that
$ (b^2)^2 = a^2 \times c^2 $

This means, even, $ a^2, b^2, c^2 $ are also in GP.

The sequence $-6 + 42 - 294 + 2058$ is a

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. finite geometric sequence

  2. finite arithmetic sequence

  3. infinite geometric sequence

  4. infinite harmonic sequence

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Correct Option: A
Explanation:

The sequence $-6 + 42 - 294 + 2058$ is a finite geometric sequence.
Here the common ratio is $-7$.

As it has finite terms, therefore the series is finite sequence.

The sum of the series $10 - 5 + 2.5 - 1.25.....$ is called

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. finite geometric sequence

  2. finite arithmetic sequence

  3. infinite geometric sequence

  4. infinite harmonic sequence

Show answer
Correct Option: C
Explanation:

Given series is $10-5+2.5-1.25.....$

Here the common ratio is $\dfrac {-5}{10}=-\dfrac {1}{2}$.
It is also never ending and continued.
Hence, the given series is infinite geometric series.

When a number $x$ is subtracted from each of the numbers $8, 16$, and $40$, the resulting three numbers form a geometric progression. Find the value of $x$.

maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

  1. $3$

  2. $4$

  3. $6$

  4. $12$

  5. $18$

Show answer
Correct Option: B
Explanation:

Given that ${(16-x)}^{2}=(8-x)(40-x)$
$\Rightarrow 256-32x+{x}^{2} = 320-48x+{x}^{2}$
$\Rightarrow 16x = 64$ 

$\Rightarrow x = 4$