Tag: maths

Questions Related to maths

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

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

  1. 6

  2. $\infty$

  3. 216

  4. 36

Show answer
Correct Option: A
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 $a = \dfrac{1}{2}, r = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} = \dfrac{1}{2}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$
$\therefore S _{\infty} = 6^1 = 6$

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

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

  1. $10$

  2. $14$

  3. $12$

  4. $22$

Show answer
Correct Option: C
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 $(2)$
$\lambda =12$

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, $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _8 = \dfrac{1(1-(2)^{8})}{1-2}$
$\Rightarrow S _8 = \dfrac{1(-255)}{-1}$
$\Rightarrow S _8 = 255$