Tag: sum to infinite terms of a gp

Questions Related to sum to infinite terms of a gp

Sum to infinity of the series $\displaystyle \frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ is

sum to infinite terms of a gp sequence, progression and series maths

  1. $\displaystyle \frac { 4 }{ 9 } $

  2. $\displaystyle \frac { 1 }{ 3 } $

  3. $\displaystyle \frac { 2 }{ 9 } $

  4. none of these

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Correct Option: C
Explanation:

Let $\displaystyle S=\frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ to $\infty$   ...(1)


Multiplying both sides by $\displaystyle -\frac { 1 }{ 2 } $, the common ratio $G.P.$


$\displaystyle -\frac { 1 }{ 2 } S=-\frac { 2 }{ 6 } +\frac { 5 }{ 12 } -\frac { 8 }{ 24 } +...$ to $\infty$    ....(2)

Subtracting (2) from (1), we have

$\displaystyle \frac { 3 }{ 2 } S=\frac { 2 }{ 3 } -\frac { 3 }{ 6 } +\frac { 3 }{ 12 } -\frac { 3 }{ 24 } +...$ to $\infty$

$\displaystyle =\frac { 2 }{ 3 } -\left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +... \right) $

$\displaystyle =\dfrac { 2 }{ 3 } -\dfrac { \dfrac { 1 }{ 2 }  }{ 1-\left( -\dfrac { 1 }{ 2 }  \right)  } =\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 } $

$\displaystyle \therefore S=\frac { 1 }{ 3 } \times \frac { 2 }{ 3 } =\frac { 2 }{ 9 } $

If $S$ is the sum to infinity of a GP, whose first term is $a$, then the sum of the first $ n$  terms is

sum to infinite terms of a gp sequence, progression and series maths

  1. $\displaystyle S\left ( 1-\frac{a}{S} \right )^{n}$

  2. $\displaystyle S\left [ 1-\left ( 1-\frac{a}{S} \right )^{n} \right ]$

  3. $\displaystyle a\left [ 1-\left ( 1-\frac{a}{S} \right )^{n} \right ]$

  4. none of these

Show answer
Correct Option: B
Explanation:

Let r be the common ration.
GIven, $S _\infty=S=\frac{a}{1-r}$
$\Rightarrow 1-r=\frac aS$
$\Rightarrow r=1-\frac aS$
Now, sum of n terms is given by
$S _n=\dfrac{a(1-r^n)}{1-r}$


       $=\dfrac{a(1-(1-\frac aS)^n)}{1-(1-\frac aS)}$

       $=\dfrac{a(1-(1-\frac aS)^n)}{\frac aS}$


       $=S[1-(1-\frac aS)^n]$
Option B is correct.

$\displaystyle2+1+\frac{1}{2}+\frac{1}{4}+\cdots\cdots\infty$ is

sum to infinite terms of a gp sequence, progression and series maths

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: D
Explanation:

The given series is a Geometric Progression, with first terms $ a =2$ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {1}{2} $

For a GP, sum to infinity is given by the formula $ \dfrac {a}{1-r} $

So, for the given series, $ S _\infty  = \dfrac {2}{1-\dfrac {1}{2}} = 4 $

What is the sum of the infinite geometric series where the beginning term is $2$ and the common ratio is $3$?

sum to infinite terms of a gp sequence, progression and series maths

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$

Show answer
Correct Option: B
Explanation:

From the given information, we have
first term $=a=2 $, common ratio $r=3$
We know $S = \dfrac{a}{1-r}$
Therefore, $S = \dfrac{2}{1-3}$
$\Rightarrow S = -1$

The value of the infinite product $6^{\frac{1}{2}}\times 6^{\frac{1}{2}}\times 6^{\frac{3}{8}}\times 6^{\frac{1}{4}}\times .........$ is

sum to infinite terms of a gp sequence, progression and series maths

  1. 6

  2. 36

  3. 216

  4. $\infty$

Show answer
Correct Option: B
Explanation:
$6^{\frac12}\times6^{\frac12}\times6^{\frac38}\times6^{\frac14}\times...............$
$=6^{\frac12+\frac24+\frac38+\frac4{16}+...................}$
$=6^{\frac12\left(1+\frac22+\frac3{2^2}+\frac4{2^3}+......................\right)}$

Let $S=1+\cfrac22+\cfrac3{2^2}+\cfrac4{2^3}+............$, then
$\cfrac S2=S-\cfrac S2$
or, $\cfrac S2=1+\cfrac12+\cfrac1{2^2}+\cfrac1{2^3}+...........$
or, $\cfrac S2=\cfrac1{1-\cfrac12}$ ..... [Using sum of infinite terms of an G.P]
or, $S=4$
Then we have,
$6^{\frac12}\times6^{\frac12}\times6^{\frac38}\times6^{\frac14}\times...............$
$=6^{\frac12\times4}=6^2=36$
Hence, B is the correct option.

Calculate the sum of the infinite series: $1 - \dfrac {1}{3} + \dfrac {1}{9} - \dfrac {1}{27} + .....$.

sum to infinite terms of a gp sequence, progression and series maths

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

  2. $\dfrac {3}{4}$

  3. $1$

  4. $\dfrac {4}{3}$

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

Show answer
Correct Option: B
Explanation:

Given series is $1,-\dfrac{1}{3},+\dfrac{1}{9},-\dfrac{1}{27}......................$

Then common ratio $=$ $ (-\dfrac{1}{3})/1=-\dfrac{1}{3}$
Then sum of infinite series $S=$ $\dfrac{a _{1}}{1-r}=\dfrac{1}{1-(-\frac{1}{3})}=\dfrac{1}{\frac{1+3}{3}}=\dfrac{3}{4}$

Calculate the sum of the infinite geometric series $2+\left(-\displaystyle\frac{1}{2}\right)+\left(\displaystyle\frac{1}{8}\right)+\left(-\displaystyle\frac{1}{32}\right)+...$

sum to infinite terms of a gp sequence, progression and series maths

  1. $1\displaystyle\frac{3}{8}$

  2. $1\displaystyle\frac{2}{5}$

  3. $1\displaystyle\frac{1}{2}$

  4. $1\displaystyle\frac{3}{5}$

  5. $1\displaystyle\frac{5}{8}$

Show answer
Correct Option: D
Explanation:

Given the geometric series is $2,\left ( -\dfrac{1}{2} \right ),\left ( \dfrac{1}{8} \right ),\left ( -\dfrac{1}{32} \right ).......................$

Then common ratio $=-\dfrac{1}{4}$
And first term is $2$.
Then sum of the infinite geometric series $=$ $S=\dfrac{a _{1}}{1-r}=\dfrac{2}{1-(-\frac{1}{4})}=\dfrac{2\times 4}{4+1}=\dfrac{8}{5}=1\dfrac{3}{8}$

The sum of first $n$ terms of an infinite G.P. is

sum to infinite terms of a gp sequence, progression and series maths

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

  2. $S _n = \dfrac{a _1(1-r^n)}{1-r}$

  3. $S = \dfrac{an}{1-r}$

  4. $S _n = \dfrac{a _1(1-r^n)}{1+r}$

Show answer
Correct Option: A
Explanation:

Sum of GP $=\dfrac{a(r^n -1)}{r-1}$

$a= $ first term
$r=$ common ratio
For $n \to \infty$
$r^n = 0$ for $r<1$
$r^n \to \infty $ for $r>1$
Thus $\text{sum} =\dfrac {a}{r-1}$

If ${S} _{p}$ denote the sum of the series $1+{r}^{p}+{r}^{2p}+..$ upto infinity and ${X} _{p}$ be the sum of the series $1-{r}^{p}+{r}^{2p}-..$ upto infinity then $\left( r\in \left( -1,1 \right) -\left{ 0 \right}  \right)$

sum to infinite terms of a gp sequence, progression and series maths

  1. ${S} _{p}+{X} _{p}={2X} _{2p}$

  2. ${S} _{p}+{X} _{p}={2S} _{2p}$

  3. ${S} _{p}+{X} _{p}={S} _{2p}$

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} { S _{ p } }\, \, is\, \, m-1+{ r^{ p } }+{ r^{ 2p } }... \ { X _{ p } }=1-{ r^{ p } }+{ r^{ 2p } } \ { S _{ p } }=\frac { 9 }{ { 1-R } } =\frac { 1 }{ { 1-{ r^{ p } } } }  \ { X _{ p } }=\frac { 1 }{ { 1+{ r^{ p } } } }  \ { S _{ p } }+{ X _{ p } }=\frac { 1 }{ { 1-{ r^{ p } } } } +\frac { 1 }{ { 1+{ r^{ p } } } }  \ =\frac { { 1+{ r^{ p } }+1 } }{ { \left( { 1-{ r^{ p } } } \right) \left( { 1+{ r^{ p } } } \right)  } } =\frac { 2 }{ { 1-{ r^{ 2p } } } }  \ =2{ S _{ p } } \  \end{array}$

The sum of an infinite geometric series whose first term is a and common ratio is r is given by

sum to infinite terms of a gp sequence, progression and series maths

  1. $\displaystyle S _{\infty} = \frac{1}{a - r}$

  2. $\displaystyle S _{\infty} = \frac{1}{r-a}$

  3. $\displaystyle S _{\infty} = \frac{a}{1 - r}$

  4. $\displaystyle S _{\infty} = \frac{1-r}{a}$

Show answer
Correct Option: C
Explanation:

Lets say the sum of infinite geometric series is:

$S _\infty=a + ar + ar^{2} + ar^{3}...... ar^{n}.....\infty$
Let us multiply r on both sides

$rS _\infty=ar+ar^{2}+.................\infty$
Let us subtract $rS _\infty$ to from $S _\infty$
So we can write

$S _\infty-rS _\infty=a$
$S _\infty(1-r)=a$

$S _\infty=\dfrac{a}{1-r}$