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

$6^{1/2}\, .\, 6^{1/4}\, .\, 6^{1/8}\, ..... \infty\, =\, ?$ 

  1. 6
  2. $\infty$

  3. 216

  4. 36
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $6^{\frac{1}{2}}.6^{\frac{1}{4}}.6^{\frac{1}{8}}....\infty$
Here power of 6 are in G.P
Sum of $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} ...\infty$

$S _{\infty} = \dfrac{a}{1-r}$
Here $a = \dfrac{1}{2}, r = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} = \dfrac{1}{2}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$
$\therefore S _{\infty} = 6^1 = 6$

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

In a geometric progression with common ratio 'q', the sum of the first 109 terms exceeds the sum of the first 100 terms by 12. If the sum of the first nine terms of the progression is $\displaystyle \frac {\lambda}{q^{100}}$ then the value of $ \lambda $ equals to

  1. $10$
  2. $14$
  3. $12$
  4. $22$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$r=q$ (common ratio)
${ S } _{ n }=\cfrac { a({ r }^{ n }-1) }{ (r-1) } \\ { S } _{ 109 }={ S } _{ 100 }+12\\ \cfrac { a({ q }^{ 109 }-1) }{ (q-1) } =\cfrac { a({ q }^{ 100 }-1) }{ (q-1) } +12\quad \quad (1)\\ \cfrac { a({ q }^{ 9 }-1) }{ (q-1) } =\cfrac { \lambda  }{ { q }^{ 100 } } \\ \lambda =\cfrac { a({ q }^{ 109 }-{ q }^{ 100 }) }{ (q-1) } \quad \quad \quad (2)$
From $(1)$ and $(2)$
$\lambda =12$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Let $\displaystyle S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ find the sum of first $20$ terms of the 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. $\displaystyle \frac{2^{19}-1}{2^{20}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } } )$

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

The $n^{th}$ term of the sequence 

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

  1. $(1000)^n$
  2. $10^{2n}$
  3. $10^{-2n}$
  4. $10^{-n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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}$

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

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

  1. $20.66$
  2. $40.66$
  3. $80.66$
  4. $100.66$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. $63.984$
  2. $32.451$
  3. $54.876$
  4. $25.458$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $\dfrac{17}{16}$
  2. $\dfrac{107}{16}$
  3. $\dfrac{117}{16}$
  4. $\dfrac{127}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
$S=\dfrac{a-ar^{n+1}}{1-r}$
$S=\dfrac{4-\frac{1}{32}}{1-\frac{1}{2}}$
$S=\dfrac{\frac{127}{32}}{\frac{1}{2}}$
$S=\dfrac{127}{16}$

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

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

  1. $178$
  2. $278$
  3. $378$
  4. $478$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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

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

  1. $1023$
  2. $2023$
  3. $3023$
  4. $4023$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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

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

  1. $256$
  2. $255$
  3. $254$
  4. $253$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _8 = \dfrac{1(1-(2)^{8})}{1-2}$
$\Rightarrow S _8 = \dfrac{1(-255)}{-1}$
$\Rightarrow S _8 = 255$