Tag: geometric sequences

Quizzes (2) Questions (161)

Questions Related to geometric sequences

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

Which of the following is a geometric series? 

  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. $3,9,18,36 , \dots$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$.

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

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

  1. $^{n+3}C _r$
  2. $^{n+1}C _{r+1}$
  3. $^nC _r$
  4. $^{(n+2)}C _r$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

  1. $\alpha \beta $
  2. $\beta \gamma $
  3. $\gamma \alpha $
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

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

  1. 1, 4, 16, 64, 244

  2. 1, 4, 24, 64, 256

  3. 1, 4, 16, 32, 256

  4. 1, 4, 16, 64, 256
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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





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

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

  1. AP

  2. GP

  3. HP

  4. NONE
Reveal answer Fill a bubble to check yourself
B Correct answer
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 

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

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

  1. $\displaystyle \frac{2}{5}$
  2. $\displaystyle \frac{3}{5}$
  3. $\displaystyle \frac{19}{27}$
  4. $\displaystyle \frac{8}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

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

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

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

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

  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{4}{3}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}}$

$= \dfrac{3}{4}$