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

If $a, b, c$ are in G.P., then

  1. $a^2, b^2, c^2$ are in G.P.
  2. $a^2(b+c), c^2 (a+b), b^2 (a+c)$ are in G.P.
  3. $\displaystyle \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in G.P.

  4. None of the above.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:  a, b, c are in G.P.

$\therefore b^{2}=ac$                                    ...eq ( 1 )
Squaring both sides:
$\Rightarrow b^{4}=a^{2}c^{2}$
$\Rightarrow (b^{2})^{2}=a^{2}c^{2}$
$\therefore a^{2},b^{2},c^{2}$ are in G.P.                    

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

Consider an infinite $G.P$. with first term $a $ and common ratio $r$, its sum is $4$ and the second term is $\dfrac {3}{4}$, then?

  1. $a=\dfrac{4}{7}, r=\dfrac{3}{7}$
  2. $a=\dfrac{3}{2}, r=\dfrac{1}{2}$
  3. $a=1, r=\dfrac{3}{4}$
  4. $a=3, r=\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Given:-

${S} _{\infty} = 4$
${a} _{2} = \cfrac{3}{4}$
$\Rightarrow ar = \cfrac{3}{4}$
$\Rightarrow 4ar = 3 ..... \left( 1 \right)$
As we know that,
${S} _{\infty} = \cfrac{a}{1 - r}$
$\therefore \cfrac{a}{1 - r} = 4$
$\Rightarrow 4r = 4 - a ..... \left( 2 \right)$
From equation $\left( 1 \right) & \left( 2 \right)$, we have
$a \left( 4 - a \right) = 3$
$\Rightarrow {a}^{2} - 4a + 3 = 0$
$\Rightarrow \left( a - 3 \right) \left( a - 1 \right) = 0$
$\Rightarrow a = 1$ or $a = 3$
Substituting the value of $a$ in equation $\left( 1 \right)$, we get
For $a = 3$
$\Rightarrow r = \cfrac{1}{4}$
For  $a = 1$
$\Rightarrow r = \cfrac{3}{4}$

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

The first term of an infinite geometric progression is x and its sum is $5$. then 

  1. $x < -10$
  2. $0 < x < 10$
  3. $-10 < x < 10$
  4. $x > 10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:-
${S} _{\infty} = 5$
$a = x$
As we know that,
${S} _{\infty} = \cfrac{a}{1 - r}$
$\therefore \cfrac{x}{1 - r} = 5$
$\Rightarrow \cfrac{x}{5} = 1 - r$
$\Rightarrow r = 1 - \cfrac{x}{5}$
Now,
$\left| r \right| < 1$
$\left| 1 - \cfrac{x}{5} \right| < 1$
$\Rightarrow -1 < 1 - \cfrac{x}{5} < 1$
$\Rightarrow -2 < \cfrac{-x}{5} < 0$
$\Rightarrow 0 < \cfrac{x}{5} < 2$
$\Rightarrow 0 < x < 10$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The first three of four given numbers are in G.P. and last three are in A.P. whose common difference is $6$. If the first and last numbers are same, then first will be?

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Last $3$ of the $4$ numbers are in AP.


Let they are, $a-d, a, a+d$. Also the first number is same as $4th, a+d$.

Therefore the $4$ numbers are $a+d, a-d, a, a+d$

The first $3$ of these are in G.P.


$\therefore (a-d)^2=a(a+d)$ But $d=6$

$\therefore (a-6)^2=a(a+6)$

$\therefore a^2-2a+36=0$

Solving the above quadratic equation, we get,

$a=2$

Therefore the series is:

$8,-4,2,8$


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


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