Tag: geometric sequences

Questions Related to geometric sequences

If the sum of infinite G.P. $p, 1, \dfrac{1}{p}, \dfrac{1}{p^2}, ......., $ is $\dfrac{9}{2}$. Then find the value of $p$.

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $1$

  2. $\dfrac{3}{2}$

  3. $3$

  4. $\dfrac{5}{2}$

Show answer
Correct Option: B,C
Explanation:

Sum of infinite series of a GP $=\dfrac {a} {1-r}$, where $a$ is the first term and $r$ is the common ratio

 
Here $a=p$ and $r=\dfrac {1}{p}$

$\Rightarrow \dfrac {p} {1-\dfrac {1}{p}}=\dfrac {9}{2}$

$\Rightarrow \dfrac {p^{2}}{p-1}=\dfrac {9}{2}$

$\Rightarrow 2p^{2}-9p+9=0$

$\Rightarrow 2p^{2}-6p-3p+9=0$

$\Rightarrow (2p-3)(p-3)=0$

$\Rightarrow p=3,\dfrac{3}{2}$    (because common ratio $r=1/p$ must be less than 1) 

The sum of 100 terms of the series .9+.09+.009.....will be:

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $1 - {\left( {\frac{1}{{10}}} \right)^{100}}$

  2. $1 + {\left( {\frac{1}{{10}}} \right)^{100}}$

  3. $1 + {\left( {\frac{1}{{100}}} \right)^{100}}$

  4. $1 - {\left( {\frac{1}{{100}}} \right)^{100}}$

Show answer
Correct Option: B

Sum $1,\sqrt { 3 } ,3......$ to $12$ terms is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $364\left( \sqrt { 3 } +1 \right)$

  2. $364\left( \sqrt { 3 } -1 \right)$

  3. $\dfrac { 364 }{ \left( \sqrt { 3 } -1 \right) } $

  4. $\dfrac { 728 }{ \left( \sqrt { 3 } +1 \right) }$

Show answer
Correct Option: A
Explanation:
$1\,,\,\,\sqrt 3 \,\,,\,3\,,\,.....12\,\,terms$
  It is a G.P with $a = 1\,,\,r\, = \sqrt 3 $
$1\,,\,\,\sqrt 3 \,\,,\,3\,,\,.....12\,\,terms$
${S _n} = \cfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
$ \Rightarrow {S _{12}} = \cfrac{{1\left( {{{\left( {\sqrt 3 } \right)}^{12}} - 1} \right)}}{{\sqrt 3  - 1}}$
$ \Rightarrow {S _{12}} = \cfrac{{{3^6} - 1}}{{\sqrt 3  - 1}} \times \cfrac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}}$
$ \Rightarrow {S _{12}} = \cfrac{{728\left( {\sqrt 3 } \right. + \left. 1 \right)}}{{3 - 1}}$
$\Rightarrow {S _{12}} = \cfrac{{728\left( {\sqrt 3 } \right. + \left. 1 \right)}}{2}$
$ \Rightarrow {S _{12}} = 364\left( {\sqrt 3 } \right. + \left. 1 \right)$

The sum of $2n$ terms of a geometric progression whose first term is $'a'$ and common ratio $'r'$ is equal to the sum of $n$ terms of a geometric progression whose first term is $'b'$ and common '$r^{2}$'. then $b$ is equal to

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. The sum of the first two terms of the first series.

  2. The sum of the first and last terms of the first series.

  3. The sum of the last two terms of the first series.

  4. None of these

Show answer
Correct Option: A
Explanation:

Given that

$\begin{array}{l} \dfrac { { a\left( { { r^{ 2n } }-1 } \right)  } }{ { r-1 } } =\dfrac { { b{ { \left( { { r^{ 2 } } } \right)  }^{ n } }-1 } }{ { { r^{ 2 } }-1 } }  \ \Rightarrow \dfrac { { a\left( { { r^{ 2n } }-1 } \right)  } }{ { r-1 } } =\dfrac { { b\left( { { r^{ 2n } }-1 } \right)  } }{ { (r-1)\left( { r+1 } \right)  } }  \ \Rightarrow b=a\left( { r+1 } \right)  \ \Rightarrow b=a+ar \end{array}$
$b$= sum of first two term of the first series.

The value of $x$ that satisfies the relation $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+.\infty$ if $|x|<1$ 

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac{-1\pm\sqrt5}{2}$

  2. $\dfrac{-1\pm3i}{2}$

  3. $0$

  4. $none$

Show answer
Correct Option: A
Explanation:
Here the first term is $a$ 
Common ratio is given by $-x$
The sum of infinite series is of an GP is given by $\dfrac{a}{1-r}\\x=dfrac{1}{1+x}\\x^2+x=1\\x^2+x-1=0$
Using quadratic formulae $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=1b=1c=-1$
$\implies x=\dfrac{-1\pm\sqrt{1+4}}{2}\\x=\dfrac{-1\pm\sqrt{5}}{2}$

In a infinite G.P. , the sum of first three terms is 70. If the extreme terms are multiplied by 4 and the middle term is multiplied by 5, the resulting terms form an A.p. then the sum to infinite terms of G.p.   

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. 120

  2. -40

  3. 160

  4. 80

Show answer
Correct Option: B
Explanation:

We have,

The three numbers be $a,ar\,and\,a{{r}^{2}}$.

Given that,


$ a+ar+a{{r}^{2}}=70 $

$ \Rightarrow a\left( 1+r+{{r}^{2}} \right)=70\,\,......\,\,\left( 1 \right) $


Also given that,

$4a,\,5ar\,and\,4a{{r}^{2}}$ in an A.P.

$\begin{align}

$ \Rightarrow 2\left( 5ar \right)=4a+4a{{r}^{2}} $

$ \Rightarrow 5r=2+2{{r}^{2}} $

$ \Rightarrow 2{{r}^{2}}-5r+2=0 $

$ \Rightarrow 2{{r}^{2}}-\left( 4+1 \right)r+2=0 $

$ \Rightarrow 2{{r}^{2}}-4r-1r+2=0 $

$ \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 $

$ \Rightarrow \left( r-2 \right)\left( 2r-1 \right)=0 $

$ \Rightarrow r-2=0,\,\,2r-1=0 $

$ \Rightarrow r=2,\,\,r=\dfrac{1}{2} $

From (1) we get,

$a=10\,\,\,at\,\,\,r=2$

And $a=40\,\,\,at\,\,\,r=\dfrac{1}{2}$

Sum of this series

$ =\dfrac{a}{1-r} $

$ =\dfrac{40}{1-2} $

$ =-40 $

This is the answer.

Find the sum of $1,\dfrac 14,\dfrac 1{16},.....$

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac 43$

  2. $\dfrac 34$

  3. $\dfrac 1 {16}$

  4. none

Show answer
Correct Option: A
Explanation:

The series form a GP with $a=1,r=\dfrac 14$

Sum of series $S _{\infty}=\dfrac{1}{1-\dfrac 14}\\dfrac 1{\dfrac 34}\\dfrac 43$

If a $ >0,  $ then the minimum value of sum of $  \dfrac{1}{a}, 1, a^{2}, a^{3}, \dfrac{1}{a^{4}}  $ is equal to

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $2$

  2. $3$

  3. $4$

  4. $5$

Show answer
Correct Option: A
Explanation:

Sum of $\dfrac{1}{a},1,a^2,a^3,\dfrac{1}{a^4}$ is $1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}$

$AM\ge GM$
$\implies \dfrac{1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}}{5}\ge \sqrt[5]{(1)(a^2)(a^3)(\frac{1}{a})(\frac{1}{a^4})}$
$1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}\ge 5\sqrt[5]{1}$

$1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}\ge 5$
The minimum value of $1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}$ is $5$

Evaluate:
$2+2^2+2^3+....+2^9=$

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $1396$

  2. $1022$

  3. $1587$

  4. $1478$

Show answer
Correct Option: B
Explanation:

$2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+......+{ 2 }^{ 9 }$

The series is in $GP$ with common difference$=\cfrac { { 2 }^{ 2 } }{ 2 } =\cfrac { { 2 }^{ 3 } }{ 2 } .....=\cfrac { { 2 }^{ 9 } }{ { 2 }^{ 8 } } =2$
Sum of $GP=\cfrac { a({ r }^{ n }-1) }{ r-1 } $ where $a$ is the first term and $r$ is the common difference and last term$=a({ r }^{ n }-1)$
So,Last term,$2.{ 2 }^{ n-1 }={ 2 }^{ 9 }\ { 2 }^{ n-1 }={ 2 }^{ 8 }\ n-1=8\ n=9$
Sum$==\cfrac { 2({ 2 }^{ 9 }-1) }{ 2-1 } =2(512-1)\ =2(511)=1022$
Answer $(B)$.

Sum of the first five terms of the geometric series $1 + \dfrac {2}{3} + \dfrac {4}{9} + $....is 

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac {211}{81}$

  2. $\dfrac {81}{211}$

  3. $-\dfrac {211}{81}$

  4. $-\dfrac {81}{211}$

Show answer
Correct Option: A
Explanation:

$\displaystyle { s } _{ 5 }=\frac { 1\times \left[ { 1-\left( { 2 }/{ 3 } \right)  }^{ 5 } \right]  }{ 1-\left( { 2 }/{ 3 } \right)  } =\frac { 211 }{ 81 } $