Mathematics · Quantitative Aptitude

Sequences and Series

230 Questions

Sequences and series involve ordered lists of numbers and the sum of their terms. The questions primarily test knowledge of arithmetic progressions, geometric progressions, and infinite series. It is an essential part of quantitative aptitude that requires strong pattern recognition skills.

Arithmetic progressionGeometric progressionInfinite seriesSum of termsNumber sequences

Sequences and Series Questions

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The series $a, ar, ar^2, ar^3, ar^4....$ is an

  1. finite geometric progression

  2. finite harmonic progression

  3. infinite geometric progression

  4. finite arithmetic progression
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$a, ar, ar^2, ar^3, ar^4....$ is an infinite geometric progression.

Here common ratio is $r$.
This can be found out as $\dfrac {ar}{a}=r, \dfrac {ar^2}{ar}=r$ and so on.
Thus the given series is in G.P.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

How many terms are there in the G.P $3,6,12,24,.........,384$?

  1. $8$
  2. $9$
  3. $10$
  4. $11$
  5. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here $a=3$ and $r=\cfrac{6}{3}=2$. Let the number of terms be $n$$.
Then, ${t}_{n}=384$ $\Rightarrow$ $a{r}^{n-1}=384$
$\Rightarrow$ $3\times {2}^{n-1}=384$
$\Rightarrow$ ${2}^{n-1}=128={2}^{7}$
$\Rightarrow$ $n-1=7$
$\Rightarrow$ $n=8$
$\therefore$ Number of terms $=8$.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The limit of the sum of an infinite number of terms in a geometric progression is $a/(1 - r)$ where a denotes the first term and $-1 <r<1$ denotes the common ratio. The limit of the sum of their squares is:

  1. $\dfrac{a^2}{(1 - r)^2}$
  2. $\dfrac{a^2}{1 + r^2}$
  3. $\dfrac{a^2}{1 - r^2}$
  4. $\dfrac{4a^2}{1 + r^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the original series is a, ar, ar^2, ..., the sum is a/(1-r). The series of squares is a^2, a^2r^2, a^2r^4, ..., which is a geometric series with first term a^2 and common ratio r^2. The sum is a^2/(1-r^2).

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Given a sequence of $4$ members, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is:

  1. $8$
  2. $16$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let for terms be $a,ar,ar^{2},a$

$\because ar,ar^{2}$ and a are in A.P

$\therefore ar^{2}-ar=6$

$\Rightarrow ar(r-1)=6$

And $a-ar=2\times 6$

$\Rightarrow a(r-1)=-12$

$\Rightarrow \dfrac{ar(r-1)}{a(r-1)}=\dfrac{-6}{12}$

$\Rightarrow r=-\dfrac{1}{2}(\because r\neq 1)$

$\therefore a(1-r)=12$

$\Rightarrow a\left(1+\dfrac{1}{2}\right)=12$

$\Rightarrow \dfrac{39}{2}=12$

$\Rightarrow a=8$

$\therefore $ Last term = $8$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Consider an infinite $G.P$. with first term $a $ and common ratio $r$, its sum is $4$ and the second term is $\dfrac {3}{4}$, then?

  1. $a=\dfrac{4}{7}, r=\dfrac{3}{7}$
  2. $a=\dfrac{3}{2}, r=\dfrac{1}{2}$
  3. $a=1, r=\dfrac{3}{4}$
  4. $a=3, r=\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Given:-

${S} _{\infty} = 4$
${a} _{2} = \cfrac{3}{4}$
$\Rightarrow ar = \cfrac{3}{4}$
$\Rightarrow 4ar = 3 ..... \left( 1 \right)$
As we know that,
${S} _{\infty} = \cfrac{a}{1 - r}$
$\therefore \cfrac{a}{1 - r} = 4$
$\Rightarrow 4r = 4 - a ..... \left( 2 \right)$
From equation $\left( 1 \right) &amp; \left( 2 \right)$, we have
$a \left( 4 - a \right) = 3$
$\Rightarrow {a}^{2} - 4a + 3 = 0$
$\Rightarrow \left( a - 3 \right) \left( a - 1 \right) = 0$
$\Rightarrow a = 1$ or $a = 3$
Substituting the value of $a$ in equation $\left( 1 \right)$, we get
For $a = 3$
$\Rightarrow r = \cfrac{1}{4}$
For  $a = 1$
$\Rightarrow r = \cfrac{3}{4}$

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The first term of an infinite geometric progression is x and its sum is $5$. then 

  1. $x < -10$
  2. $0 < x < 10$
  3. $-10 < x < 10$
  4. $x > 10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:-
${S} _{\infty} = 5$
$a = x$
As we know that,
${S} _{\infty} = \cfrac{a}{1 - r}$
$\therefore \cfrac{x}{1 - r} = 5$
$\Rightarrow \cfrac{x}{5} = 1 - r$
$\Rightarrow r = 1 - \cfrac{x}{5}$
Now,
$\left| r \right| < 1$
$\left| 1 - \cfrac{x}{5} \right| < 1$
$\Rightarrow -1 < 1 - \cfrac{x}{5} < 1$
$\Rightarrow -2 < \cfrac{-x}{5} < 0$
$\Rightarrow 0 < \cfrac{x}{5} < 2$
$\Rightarrow 0 < x < 10$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The third term of a geometric progression is $4$. The product of the first five terms is 

  1. ${4}^{3}$
  2. ${4}^{4}$
  3. ${4}^{5}$
  4. ${4}^{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $a$ and $r$ the first term and common ratio, respectively.
Given:Third term$=a{r}^{2}=4$
Product of first $5$ terms is
$=a.ar.a{r}^{2}.a{r}^{3}.a{r}^{4}$
$={a}^{5}{r}^{10}$
$={\left(a{r}^{2}\right)}^{5}$
$={4}^{5}$
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

In a GP the sum of three numbers is $14 ,$ if $1$ is added to first two numbers and the third number is decreased by $1$, the series becomes AP, find the geometric sequence.

  1. $2,4,8$
  2. $8,4,2$
  3. $6,18,54$
  4. $8,16,32$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation


$\\Let\>the\>numbers\>an\>a,\>ar,\>ar^2\\where\>a=first\>term\>and\>r=common\>ratio\\\therefore\>a+ar+ar^2=14\\and\\a+1,\>ar+1,\>ar^2-1\>are\>in\>AP\\\therefore\>2(ar+1)=(a+1)+(ar^2-1)\\or\>2ar+2=a+ar^2\\or\>3ar+2=a+ar+ar^2\\or\>3ar+2=14\\\therefore\>ar=4\\or\>a=(\frac{4}{r})\\\therefore(\frac{4}{r})+(\frac{4}{r})\times\>r+(\frac{4}{r})\times\>r^2=14\\or\>(\frac{4}{r})+4+4r=14\\or\>4+4r+4r^2=14r\\or\>4r^2-10r+4=0\\or\>4r^2-8r-2r+4=0\\or\>4r(r-2)-2(r-2)=0\\or\>(4r-2)(r-2)=0\\\therefore\>r=(\frac{1}{2})\>or\>2\\ifr=(\frac{1}{2}),then\>a=(\frac{4}{(\frac{1}{2})})=8\\\therefore\>sequence\>8,4,2\>\\\>and\>if\>r=2,\>then\>a=(\frac{4}{2})=2\\\therefore\>sequence\>2,4,8$

 

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Write down the first five terms of the geometric progression which has first term 1 and common ratio 4.

  1. 1, 4, 16, 64, 244

  2. 1, 4, 24, 64, 256

  3. 1, 4, 16, 32, 256

  4. 1, 4, 16, 64, 256
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let a and d be the first term and common ratio of the GP respectively.
Given a=1 and d=4.
Now, $a _n=ar^{n-1}$
$\therefore a _1=a=1$
$a _2=ar=1\times4=4$
$a _3=ar^2=1\times(4)^2=16$
$a _4=ar^3=1\times(4)^3=64$
$a _5=ar^4=1\times(4)^4=256$





Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Find the sum the infinite G.P.: $\displaystyle {\frac{2}{3}\, -\, \frac{4}{9}\, +\, \frac{8}{27}\, -\, \frac{16}{21}\, +\, ........}$ 

  1. $\displaystyle \frac{2}{5}$
  2. $\displaystyle \frac{3}{5}$
  3. $\displaystyle \frac{19}{27}$
  4. $\displaystyle \frac{8}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=\dfrac{2}{3} ,r=-\dfrac{2}{3}$
$\therefore S _{\infty }=\dfrac{\frac{2}{3}}{1-\left ( -\frac{2}{3} \right )}$


$\Rightarrow \dfrac{\frac{2}{3}}{1+\frac{2}{3}}$

$\Rightarrow \dfrac{\frac{2}{3}}{\frac{5}{3}}$

$\Rightarrow \dfrac{2}{5}$

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The sum of the series $10 - 5 + 2.5 - 1.25.....$ is called

  1. finite geometric sequence

  2. finite arithmetic sequence

  3. infinite geometric sequence

  4. infinite harmonic sequence
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given series is $10-5+2.5-1.25.....$

Here the common ratio is $\dfrac {-5}{10}=-\dfrac {1}{2}$.
It is also never ending and continued.
Hence, the given series is infinite geometric series.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

For the infinite series $1-\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } -\cfrac { 1 }{ 16 } -\cfrac { 1 }{ 32 } +\cfrac { 1 }{ 54 } -\cfrac { 1 }{ 128 } -....\quad $ let $S$ be the (limiting) sum. Then $S$ equals

  1. $0$
  2. $\cfrac { 2 }{ 7 } $
  3. $\cfrac { 6 }{ 7 } $
  4. $\cfrac { 9 }{ 32 } $
  5. $\cfrac { 27 }{ 32 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Combine the terms in threes, to get the geometric series
$\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 32 } +\cfrac { 1 }{ 256 } +....;\quad \quad S=\cfrac { \cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 8 }  } =\cfrac { 2 }{ 7 } $ or
rearrange the terms into three series:
$1+\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 64 } +...\quad -\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 16 } -\cfrac { 1 }{ 128 } -....,\quad -\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 32 } -\cfrac { 1 }{ 256 } -....\quad $
${ S } _{ 1 }=\cfrac { 1 }{ 1-\cfrac { 1 }{ 8 }  } =\cfrac { 8 }{ 7 } ;{ S } _{ 2 }=\cfrac { -\cfrac { 1 }{ 2 }  }{ 1-\cfrac { 1 }{ 8 }  } =-\cfrac { 4 }{ 7 } ;{ S } _{ 3}=\cfrac { -\cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 8 }  } =-\cfrac { 2 }{ 7 } ;\quad \therefore S=\cfrac { 2 }{ 7 } $

Multiple choice

What is the formula for calculating the future value of a single sum?

  1. FV = PV * (1 + r)^n

  2. FV = PV * (1 - r)^n

  3. FV = PV * r^n

  4. FV = PV / (1 + r)^n
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The formula for calculating the future value (FV) of a single sum is FV = PV * (1 + r)^n, where PV is the present value, r is the interest rate, and n is the number of compounding periods.

Multiple choice

What is the formula for calculating the present value of a single sum?

  1. PV = FV / (1 + r)^n

  2. PV = FV * (1 + r)^n

  3. PV = FV * r^n

  4. PV = FV - r^n
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The formula for calculating the present value (PV) of a single sum is PV = FV / (1 + r)^n, where FV is the future value, r is the interest rate, and n is the number of compounding periods.

Multiple choice

What are the first few terms of the Fibonacci sequence?

  1. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

  2. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

  3. 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...

  4. 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The first few terms of the Fibonacci sequence are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...