Tag: sum of terms of g.p

Questions Related to sum of terms of g.p

Let $\displaystyle S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ find the sum of first $20$ terms of the series

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

  1. $\displaystyle \frac{2^{20}-1}{2^{20}}$

  2. $\displaystyle \frac{2^{19}-1}{2^{19}}$

  3. $\displaystyle \frac{2^{20}-1}{2^{19}}$

  4. $\displaystyle \frac{2^{19}-1}{2^{20}}$

Show answer
Correct Option: C
Explanation:

$S=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } .......$ first $20$ terms

$n=20$ and series is in $GP$ with common difference $=\cfrac { \cfrac { 1 }{ 2 }  }{ 1 } =\cfrac { \cfrac { 1 }{ 4 }  }{ \cfrac { 1 }{ 2 }  } =\cfrac { 1 }{ 2 } $
$ a=1\quad r=\cfrac { 1 }{ 2 } $
Sum$=\cfrac { a(1-{ r }^{ n }) }{ 1-r } $  when$\quad r<1$
$ =\cfrac { 1(1-{ (\cfrac { 1 }{ 2 } ) }^{ 20 }) }{ 1-\cfrac { 1 }{ 2 }  } \ =\cfrac { (1-\cfrac { 1 }{ { 2 }^{ 20 } } ) }{ \cfrac { 1 }{ 2 }  } \ =2(1-\cfrac { 1 }{ { 2 }^{ 20 } } )\ =\cfrac { 2({ 2 }^{ 20 }-1) }{ { 2 }^{ 20 } } \ =(\cfrac { { 2 }^{ 20 }-1 }{ { 2 }^{ 19 } } )$

The $n^{th}$ term of the sequence 

$\displaystyle\frac{1}{100}$, $\displaystyle\frac{1}{10000}$, $\displaystyle\frac{1}{1000000}$, $\dots\dots$ is

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

  1. $(1000)^n$

  2. $10^{2n}$

  3. $10^{-2n}$

  4. $10^{-n}$

Show answer
Correct Option: C
Explanation:

The given series is a Geometric Progression, with first terms $ a = \dfrac {1}{100} $ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {\dfrac {1}{10000}}{\dfrac {1}{100}} = \dfrac {1}{100} $

For a GP, the $ nth $ term is given by $ T _n = ar^{n-1} =\dfrac {1}{100}  \times (\dfrac {1}{100})^{n-1} =(\dfrac {1}{100})^{n} = 10^{-2n}$

Find $S _n$, the sum of the first $n$ terms, for the following geometric series. $a _1=120, a _5= 1, r=-2$.

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

  1. $20.66$

  2. $40.66$

  3. $80.66$

  4. $100.66$

Show answer
Correct Option: B
Explanation:

Given, first term, $a = 120$, common ratio, $r = -2$ and $a _5=1$
We know $S _n=\dfrac{a _1-a _nr}{1-r}$
$S _n=\dfrac{120-(-2)}{1-(-2)}$
$S _n=\dfrac{120-(-2)}{1-(-2)}$
$S _n=\dfrac{122}{3}$
$S _n=40.66$

Find the sum of the first $6$ terms of the geometric series $80 - 20 + 5 +.....$

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

  1. $63.984$

  2. $32.451$

  3. $54.876$

  4. $25.458$

Show answer
Correct Option: A
Explanation:

First term, $a$ is $80$
Common ratio, $r =$ $\dfrac{-20}{80}=\dfrac{-1}{4}$
$S _n=\dfrac{a(1-r^2)}{1-r}$
$S _n=\dfrac{80(1-(\frac{-1}{4})^2)}{1-\frac{-1}{4}}$
$S _n = \dfrac{79.98}{1.25}$
$S _n = 63.98$

Find the sum of the geometric series $4 + 2 + 1 +... +$ $\dfrac{1}{16}$

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

  1. $\dfrac{17}{16}$

  2. $\dfrac{107}{16}$

  3. $\dfrac{117}{16}$

  4. $\dfrac{127}{16}$

Show answer
Correct Option: D
Explanation:

Given series is $4+2+1+.....+\dfrac {1}{16}$
First term, $a$ is $4$
Common ratio, $r =$ $\dfrac{2}{4}=\dfrac{1}{2}$
Use the formula for the sum of the geometric series.
$ar^n$ is a next term.
$\dfrac{1}{16}=\dfrac{1}{2}\times \dfrac{1}{16}=\dfrac{1}{32}$ is the next term.
$S=\dfrac{a-ar^{n+1}}{1-r}$
$S=\dfrac{4-\frac{1}{32}}{1-\frac{1}{2}}$
$S=\dfrac{\frac{127}{32}}{\frac{1}{2}}$
$S=\dfrac{127}{16}$

What is $S _6$ of the geometric progression $6, 12, 24...$?

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

  1. $178$

  2. $278$

  3. $378$

  4. $478$

Show answer
Correct Option: C
Explanation:

Given series is $6,112,24,....$
To find the sum of the first $S _n$ terms of a geometric sequence by using the formula,
Here $a = 6, r = 2, n = 6$
$S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _6 = \dfrac{6(1-(2)^6)}{1-2}$
$ = \dfrac{6(-63)}{-1}$
$ = 378$

Find $3 + 12 + 48 +...$ up to $5$ terms.

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

  1. $1023$

  2. $2023$

  3. $3023$

  4. $4023$

Show answer
Correct Option: A
Explanation:

Given series is $3+12+48+....$ upto $5$ terms
To find the sum of the first $S _n$ terms of a geometric sequence by using the formula,
Here $a = 3, r = 4, n = 5$
$S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _5 = \dfrac{3(1-(4)^{5})}{1-4}$
$ = \dfrac{3(-1023)}{-3}$
$ = 1023$

Determine the sum of the first 8 terms of the G.P. $1, 2, 4, 8...$

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

  1. $256$

  2. $255$

  3. $254$

  4. $253$

Show answer
Correct Option: B
Explanation:
Given series is $1,2,4,8,....$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula.
From given question, we have
$a = 1, r = 2, n = 8$
Therefore, $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _8 = \dfrac{1(1-(2)^{8})}{1-2}$
$\Rightarrow S _8 = \dfrac{1(-255)}{-1}$
$\Rightarrow S _8 = 255$

Evaluate the sum of the first nine terms of the geometric sequence $5, 10, 20,...$

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

  1. $1555$

  2. $2555$

  3. $3555$

  4. $4555$

Show answer
Correct Option: B
Explanation:

Given sequence is $5,10,20,....$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula
Here $a = 5, r = 2, n = 9$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _9 = \dfrac{5(1-2^{9})}{1-2}$
$\Rightarrow S _9 = \dfrac{-2555}{-1}$
$\Rightarrow S _9 = 2555$

Calculate the sum of first $20$ terms of the G.P. $-1, 1, -1, 1....$

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: A
Explanation:
Given series is $-1,1,-1,1,....$
Here $a = -1, r = -1$
We know the formula, $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{20} = \dfrac{1(1-(-1)^{20})}{1-(-1)}$
$ = \dfrac{1(1-(-1)^{20})}{2}$
$= 0$