Tag: geometric sequences

Questions Related to geometric sequences

Which of the following is a geometric series? 

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

  1. $2,4,6,8 , \dots \dots$

  2. $1 / 2,1,2,4 \dots \dots$

  3. $1 / 4,1 / 6,1 / 8,1 / 10 , \dots \ldots$

  4. $3,9,18,36 , \dots$

Show answer
Correct Option: B
Explanation:

A series is said to be geometric when ratio between one term and its previous term is constant/same throughout the series.


a) $2, 4, 6, 8$


$\Rightarrow$$\dfrac{4}{2} \neq \dfrac{6}{4} $

b) $\dfrac{1}{2} , 1 , 2, 4$

$\Rightarrow$$\dfrac{1}{\dfrac{1}{2}} = \dfrac{2}{1} = \dfrac{4}{2} = 2 $

c) $\dfrac{1}{4} , \dfrac{1}{6} , \dfrac{1}{8}$

$\Rightarrow$$\dfrac{1/6}{1/4} \neq \dfrac{1/8}{1/6}$

d) $3, 9, 18 , 36$

$\Rightarrow$$\dfrac{9}{3} \neq \dfrac{18}{9}$.

Coefficient of $x^r$ in $1+(1+x)+(1+x)^2+......+ (1+x)^n$ is 

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

  1. $^{n+3}C _r$

  2. $^{n+1}C _{r+1}$

  3. $^nC _r$

  4. $^{(n+2)}C _r$

Show answer
Correct Option: A
Explanation:

Consider given the series, $1+\left( 1+x \right)+{{\left( 1+x \right)}^{2}}+{{\left( 1+x \right)}^{3}}+...............{{\left( 1+x \right)}^{n}}$

Given series is G.P. which have first term $a=1$,comman ratio $r=1+x$ and number of term $=n$

So sum is,

${{s} _{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{[{{\left( 1+x \right)}^{n}}-1]}{1+x}={{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$

Now rth term of ${{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$ is $={}^{n-1}{{C} _{r}}{{x}^{r}}-{}^{-n}{{C} _{r}}.{{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right){{x}^{r}}$


Hence, coefficient of ${{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right)$


Hence, this is the answer.

The value of $p$ if $3,p,12$ are in GP

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

  1. $6$

  2. $4$

  3. $9$

  4. None.

Show answer
Correct Option: A
Explanation:

Given $3,p,12$ are in GP

The condition to be in GP
$b^2=ac\\p^2=3(12)\\p^2=36\\p=6$

The common ratio of GP $4,8,16,32,.....$ is

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

  1. $2$

  2. $3$

  3. $4$

  4. $0$

Show answer
Correct Option: A
Explanation:

Given series is $4,8,16,32,....$

Common ratio is given as $\dfrac84=2$

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. $0$

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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}}$

$= \dfrac{3}{4}$