Quantitative Aptitude

Number Series

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Number Series Questions

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

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

  1. $\dfrac {211}{81}$
  2. $\dfrac {81}{211}$
  3. $-\dfrac {211}{81}$
  4. $-\dfrac {81}{211}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

The sum of sequence $0.15,0.015,0.0015,.....$ upto 20 term is ?

  1. $\dfrac{1}{6}[1-(0.1)^{20}]$
  2. $\dfrac{1}{6}[1+(0.1)^{20}]$
  3. $\dfrac{1}{3}[1-(0.1)^{20}]$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the sequence in Geometric progression 

where, $a=0.15;\ r=\dfrac { 0.015 }{ 0.15 } =0.1;\ n=20\ \therefore { S } _{ n }=\dfrac { a({ r }^{ n }-1) }{ (r-1) } \ =\dfrac { 0.15({ \left( 0.1 \right)  }^{ 20 }-1) }{ 0.1-1 } \ =\dfrac { 1 }{ 6 } \left( 1-{ (0.1) }^{ 20 } \right) $

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

The sum of first $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+...$ is

  1. $121(\sqrt{6}+\sqrt{2})$
  2. $243(\sqrt{3}+1)$
  3. $\cfrac{121}{\sqrt{3}-1}$
  4. $242(\sqrt{3}-1)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$s=\sqrt{2}+\sqrt{6}+\sqrt{18}+....$

$\Rightarrow s=\sqrt{2}+\sqrt{2}\times \sqrt{3}+\sqrt{2}(\sqrt{3})^2+.....$

This is a G.P with $1$st term $(a)=\sqrt{2}$ and common ratio$(r)=\sqrt{3}$

Sum of $10$ terms of this G.P., $S=\dfrac{a(r^{10}-1)}{r-1}$

$=\dfrac{\sqrt{2}((\sqrt{3})^{10}-1)}{\sqrt{3}-1}$

$=\dfrac{\sqrt{2}(242)}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

$=121\times \sqrt{2}(\sqrt{3}+1)$

$=121(\sqrt{6}+\sqrt{2})$

$\Rightarrow (A)$ Option. 

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

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

The sum of $6^{th}$ term in the geometric series $4, 12, 36...$ is

  1. $1456$
  2. $2456$
  3. $3456$
  4. $4456$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given sequence is $4,12, 36$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula
Here $a = 4, r = 3, n = 6$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _6 = \dfrac{4(1-3^{6})}{1-3}$
$\Rightarrow S _6 = \dfrac{-2912}{-2}$
$\Rightarrow S _6 = 1456$

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

What is the sum of first eight terms of the series $1-\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 8 } +.....$?

  1. $\cfrac { 89 }{ 128 } $
  2. $\cfrac { 57 }{ 384 } $
  3. $\cfrac { 85 }{ 128 } $

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given series is a sum of terms of GP with common ratio $-\dfrac{1}{2}$
Sum of $n$ terms of GP with common ratio $r$ and first term $a$ is $\dfrac { a(1-r^{ n }) }{ 1-r } $
Putting $a=1,n=8$ and $r=-\dfrac { 1 }{ 2 } $ in above equation, we have 

Sum $=\dfrac { 1(1-(-\frac { 1 }{ 2 } )^{ 8 }) }{ 1-(-\frac { 1 }{ 2 } ) } =\dfrac { 1-\frac { 1 }{ 256 }  }{ \frac { 3 }{ 2 }  } =\dfrac { 255 }{ 128\times 3 } =\dfrac { 85}{128} $
Hence, option C is correct

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

The sum of $2n$ terms of a series of which every even term is $'a'$ times the terms before it, and every odd term $'c'$ times the terms before it, the first term being unity, is

  1. $\dfrac { \left( 1-a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$
  2. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac+1 }$
  3. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { T _{ 1 } }=1 \ { T _{ 2 } }=a \ { T _{ 3 } }=Ca \ { T _{ 4 } }=C{ a^{ 2 } } \ { T _{ 2n } }=a\frac { { 2n } }{ 2 } \cdot C\frac { { 2n } }{ 2 } -1={ a^{ n } }{ C^{ n-1 } } \ { 5 _{ 2n } }=1+a+ca.....{ a^{ n } }{ C^{ n-1 } } \ =1+\left[ { a+c{ a^{ 2 } }+{ c^{ 2 } }{ a^{ 3 } }...{ a^{ n } }{ c^{ n-1 } } } \right]  \ +\left[ { ca+{ c^{ 2 } }{ a^{ 2 } }+{ c^{ 3 } }{ a^{ 3 } }.....{ c^{ n-1 } }{ a^{ n-1 } } } \right]  \ =1+\frac { { a\left( { { a^{ n } }{ c^{ n-1 } } } \right)  } }{ { ac-1 } } +\frac { { ac\left( { { a^{ n-1 } }{ c^{ n-1 } }-1 } \right)  } }{ { ac-1 } }  \ =\frac { { \left( { { a^{ n } }{ c^{ n } }-1 } \right) \left( { a+1 } \right)  } }{ { ac-1 } }  \end{array}$

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

The sum of $10$ terms of the series $0.7 + .77 + .777 + \ldots \ldots \ldots$ is

  1. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$
  2. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$
  3. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$
  4. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$0.7+0.77+0.777+......$
$=7\left( 0.1+0.11+0.111+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 0.9+0.99+0.999+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 1-0.1+1-0.1+1-0.001+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\left( 0.1+0.01+0.001+...... \right)  \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( 1-{ 10 }^{ -10 } \right)  }{ 1-0.1 }  \right) =\frac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( { 10 }^{ 10 }-1 \right)  }{ 0.9\times { 10 }^{ 10 } }  \right) =\dfrac { 7 }{ 9 } \left( 10-\dfrac { 1 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 9 } \left( \dfrac { 89 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 81 } \left( 89+\dfrac { 1 }{ { 10 }^{ 10 } }  \right) $      [B]
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of series $\displaystyle \frac{3}{4} + \frac{15}{16} + \frac{63}{64}+ ..... $ up to $n$ terms is

  1. $\displaystyle n - \frac{4^n}{3} - \frac{1}{3}$
  2. $\displaystyle n + \frac{4^{-n}}{3} - \frac{1}{3}$
  3. $\displaystyle n + \frac{4^n}{3} - \frac{1}{3}$
  4. $\displaystyle n - \frac{4^{-n}}{3} - \frac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For $n=1$, we have
$\displaystyle n - \dfrac{4^n }{3} - \dfrac{1}{3} = 1  - \dfrac{4}{3} - \dfrac{1}{3} = - \dfrac{2}{3}$
$\displaystyle n + \dfrac{4^n}{3} - \dfrac{1}{3} = 1 + \dfrac{4}{3} - \dfrac{1}{3} = 2$
$n - \displaystyle \dfrac{4^{-n}}{3} + \dfrac{1}{3} = 1 - \dfrac{4^{-1}}{3} + \dfrac{1}{3}= \dfrac{5}{4}$
Also, for $n = 2$, we have
$ \displaystyle n + \dfrac{4^{-n}}{3} - \dfrac{1}{3} = 2 + \dfrac{1}{48} - \dfrac{1}{3} = \dfrac{27}{16}$ and $\displaystyle \dfrac{3}{4} + \dfrac{15}{16} = \dfrac{27}{16}$
Hence, option (b) is correct.
ALTER We have,
$\displaystyle \dfrac{3}{4} + \dfrac{15}{16} + \dfrac{63}{64}+ ..... $ to n terms
$= \displaystyle \dfrac{2^2 - 1}{2^2} + \dfrac{2^4 - 1}{2^4} + \dfrac{2^6 - 1}{2^6}+ .... $ to n terms.
$= \displaystyle \left ( 1 - \dfrac{1}{2^2} \right ) + \left ( 1 - \dfrac{1}{2^4} \right ) + \left( 1 - \dfrac{1}{2^6} \right ) + ..... $ to n terms
$= n - \left \{ \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^6} + .... \text{to n terms} \right \}$
$= n \displaystyle - \dfrac{1}{2^2} \left \{ \dfrac{1 - \left (\dfrac{1}{2^2} \right )^n }{1 - \dfrac{1}{2^2}} \right \}$
$= \displaystyle n - \dfrac{1}{3} (1 - 4^{-n})$
$= n + \displaystyle \dfrac{4^{-n}}{3} - \dfrac{1}{3}$

Multiple choice maths average arithmetic mean of ap introduction to averages means

What is the average of the first $300$ terms of the given sequence?
$1, -2, 3, -4, 5, -6, ....., n.(-1)^{n + 1}$

  1. $-1$
  2. $0.5$
  3. $0$
  4. $-0.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Avg $=\cfrac{[1+3+5+7+...+(2n-1)]-[2+4+6+...+2n]}{n}$
Here $n=300$
No. of even terms $=150$
No. of odd terms $=150$
Now,
$1+3+5+7+...+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n1}=\cfrac{150}{2} [2\times 1 +149\times 2]=22500$
and $2+4+6+8+....+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n2}=\cfrac{150}{2}[2\times 2+149\times 2]=22650$
Average $=\cfrac{S _{n1}-S _{n2}}{300}=\cfrac{22500-22650}{300} \\ =-0.5$