Tag: introduction to geometric progression

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

The sum of $1 + \left( {1 + a} \right)x + \left( {1 + a + {a^2}} \right){x^2} + ....\infty ,\,0 < a,\,x < 1$ equals

  1. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 - a} \right)}}$
  2. $\dfrac{1}{{\left( {1 - a} \right)\left( {1 - ax} \right)}}$
  3. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 - ax} \right)}}$
  4. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 + a} \right)}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$1+\left( 1+a \right) x+\left( 1+a+{ a }^{ 2 } \right) { x }^{ 2 }+\dots \infty $

$ General\quad term\quad of\quad series={ T } _{ n }=\frac { \left( 1-{ a }^{ r+1 } \right)  }{ \left( 1-a \right)  } { x }^{ r }$  

$ { S } _{ n }=\sum _{ r=1 }^{ \infty  }{ \dfrac { \left( 1-{ a }^{ r+1 } \right) { x }^{ r } }{ \left( 1-a \right)  }  } $  

$ \left( 1-a \right) { S } _{ n }=\sum _{ r=1 }^{ \infty  }{ { x }^{ r } } -a\sum _{ r=1 }^{ \infty  }{ \left( ax \right) ^{ r } } $ 

$\left( 1-a \right) { S } _{ n }=\dfrac { 1 }{ 1-x } -\dfrac { a }{ 1-ax } $     ...........  $\because G.P.$ series 

$ \left( 1-a \right) { S } _{ n }=\dfrac { 1-ax-a+ax }{ \left( 1-x \right) \left( 1-ax \right)  } $  

$ \left( 1-a \right) { S } _{ n }=\dfrac { 1-a }{ \left( 1-x \right) \left( 1-ax \right)  } $  

$ { S } _{ n }=\dfrac { 1 }{ \left( 1-x \right) \left( 1-ax \right)  } $
B is correct
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If roots of the equations $(b-c)x^2+(c-a)x+a-b=0$, where $b\neq c$, are equal, then a, b, c are in?

  1. G.P.

  2. H.P.

  3. A.P.

  4. A.G.P.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(b-c)x^{2}+(c-a)x+a-b=0$
Root are equal, so $D=0$
$\Rightarrow (c-a)^{2}-4(b-c)(a-b)=0$
$\Rightarrow c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0$
$\Rightarrow c^2+a^2+4b^2+2ac-4ab-4bc=0$
$\Rightarrow (a+c-2b)^{2}=0$
$a+c=2b$

























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

If the roots of ${ x }^{ 2 }-k{ x }^{ 2 }+14x-8=0$ are in geometric progression, then $k=$

  1. $-3$
  2. $7$
  3. $4$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow \frac { a }{ r } ,a,ar=8\quad \quad \Rightarrow { a }^{ 3 }=8\quad \quad \Rightarrow a=2$
$a=2$ is a root of the given equation 
$\Rightarrow 8-4k+28-8=0\quad \quad \Rightarrow k=7$

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

The third term of a geometric progression is $4$. The product of the first five terms is 

  1. ${4}^{3}$
  2. ${4}^{4}$
  3. ${4}^{5}$
  4. ${4}^{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $a$ and $r$ the first term and common ratio, respectively.
Given:Third term$=a{r}^{2}=4$
Product of first $5$ terms is
$=a.ar.a{r}^{2}.a{r}^{3}.a{r}^{4}$
$={a}^{5}{r}^{10}$
$={\left(a{r}^{2}\right)}^{5}$
$={4}^{5}$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

In a GP the sum of three numbers is $14 ,$ if $1$ is added to first two numbers and the third number is decreased by $1$, the series becomes AP, find the geometric sequence.

  1. $2,4,8$
  2. $8,4,2$
  3. $6,18,54$
  4. $8,16,32$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation


$\\Let\>the\>numbers\>an\>a,\>ar,\>ar^2\\where\>a=first\>term\>and\>r=common\>ratio\\\therefore\>a+ar+ar^2=14\\and\\a+1,\>ar+1,\>ar^2-1\>are\>in\>AP\\\therefore\>2(ar+1)=(a+1)+(ar^2-1)\\or\>2ar+2=a+ar^2\\or\>3ar+2=a+ar+ar^2\\or\>3ar+2=14\\\therefore\>ar=4\\or\>a=(\frac{4}{r})\\\therefore(\frac{4}{r})+(\frac{4}{r})\times\>r+(\frac{4}{r})\times\>r^2=14\\or\>(\frac{4}{r})+4+4r=14\\or\>4+4r+4r^2=14r\\or\>4r^2-10r+4=0\\or\>4r^2-8r-2r+4=0\\or\>4r(r-2)-2(r-2)=0\\or\>(4r-2)(r-2)=0\\\therefore\>r=(\frac{1}{2})\>or\>2\\ifr=(\frac{1}{2}),then\>a=(\frac{4}{(\frac{1}{2})})=8\\\therefore\>sequence\>8,4,2\>\\\>and\>if\>r=2,\>then\>a=(\frac{4}{2})=2\\\therefore\>sequence\>2,4,8$

 

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.