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 arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If ${ a } _{ 1 },{ a } _{ 2 },{ a } _{ 3 },\dots $ are terms of AP such that ${ a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$, then the sum of first $24$ terms is

  1. $9\times { 10 }^{ 2 }$
  2. $9\times { 10 }^{ 3 }$
  3. $10\times { 9 }^{ 2 }$
  4. $10\times { 9 }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that the sum of terms of AP equidistant from the beginning and end is always same and it is always equal to the sum of first and last terms.
$\Rightarrow { a } _{ 1 }+{ a } _{ 24 }={ a } _{ 6 }+{ a } _{ 20 }={ a } _{ 10 }+{ a } _{ 15 }$
$\because { a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$
$\therefore 3\left( { a } _{ 1 }+{ a } _{ 24 } \right) =225\Rightarrow { a } _{ 1 }+{ a } _{ 24 }=75$
$\therefore { S } _{ 24 }=\dfrac { 24 }{ 2 } \left( { a } _{ 1 }+{ a } _{ 24 } \right)$   $\left[ \because { S } _{ n }=\dfrac { n }{ 2 } \left( { a } _{ 1 }+{ a } _{ n } \right)  \right] $
           $=12\left( 75 \right) =900=9\times { 10 }^{ 2 }$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If the $n^{th}$ term of an AP be $(2n-1)$, then the sum of its first n terms will be.

  1. $n^2-1$
  2. $(2n-1)^2$
  3. $n^2$
  4. $n^2+1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$a _n=(2n-1)$
$\Rightarrow$  $a _1=2\times 1-1=1$
$\Rightarrow$  $a _2=2\times 2-1$
            $=4-1$
            $=3$
$\Rightarrow$  $d=a _1-a _1=3-1$
$\therefore$  $d=2$
$\Rightarrow$  $S _1=\dfrac{n}{2}[2a _1+(n-1)d]$
           
            $=\dfrac{n}{2}[2(1)+(n-1)2]$

            $=\dfrac{n}{2}[2+2n-2]$

            $=\dfrac{n}{2}\times 2n$

            $=n^2$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

Find the sum of the first $15$ terms of the following sequences having $n$th term as
${a} _{n}=3+4n$

  1. 525

  2. 563

  3. 184

  4. 189
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$a _{n}=3+4n$ (Given)

Now, 
$a _{1}=3+4\times 1=7$
$a _{2}=3+4\times 2=11$
$a _{3}=3+4\times 3=15$
So the series is
The sum of first is turns is
$S _{n}=\dfrac{n}{2}[2a+(n-1)d]$
$a=7, n=15, d=4$
$S _{n}=\dfrac{15}{2}[2\times 7+(15-1).4]$
$S _{n}=\dfrac{15}{2}[14+56]$
$S _{n}=\dfrac{15\times 70}{2}$
$S _{n}=15\times 35$
$S _{n}=525$
The sum of first $15$ terms of given series is 
$S _{n}=525$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

Let ${V} _{r}$ denote the sum of the first $r$ terms of an A.P whose first term is $r$ and common difference is $(2r-1)$.Let

${T} _{r}={V} _{r+1}-{V} _{r}-2$ and 

${Q} _{r}={T} _{r+1}-{T} _{r}$ $T$ is always

  1. an odd number

  2. an even number

  3. a prime number

  4. a composite number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have sum of $n$ terms $=\dfrac{n}{2}\left(2a+(n-1)d\right)$ where $a$ is the first term, $n$ is the number of terms and $d$ is the common difference in an A.P.
From the passage $n=r,$ $a=2r$ and $d=(2r-1)$
$\therefore {V} _{r}=\dfrac{r}{2}\left[2r+(r-1)(2r-1)\right]$
$\Rightarrow \dfrac{r}{2}\left[2r+2{r}^{2}-3r+1\right]=\frac{r}{2}\left[2{r}^{2}-r+1\right]$
Thus ${V} _{r}=\frac{1}{2}\left[2{r}^{3}-{r}^{2}+r\right]={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}$
Now ${T} _{r}={V} _{r+1}-{V} _{r}-2$
From above ${V} _{r}={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}$ 
${V} _{r+1}={\left(r+1\right)}^{3}-\frac{{\left(r+1\right)}^{2}}{2}+\frac{\left(r+1\right)}{2}$
We have ${T} _{r}={V} _{r+1}-{V} _{r}-2$
${T} _{r}={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}-\left({\left(r+1\right)}^{3}-\frac{{\left(r+1\right)}^{2}}{2}+\frac{\left(r+1\right)}{2}\right)-2$
On simplifying, we get

${T} _{r}={\left(r+1\right)}^{3}-{r}^{3}-\frac{1}{2}\left({\left(r+1\right)}^{2}-{r}^{2}\right)+\frac{1}{2}\left(r+1-r\right)-2$
$\Rightarrow{T} _{r}=\left(r+1-r\right)\left({\left(r+1\right)}^{2}+r\left(r+1\right)+{r}^{2}\right)+\frac{1}{2}-2$
On simplifying, we get
${T} _{r}={r}^{2}+1+2r+{r}^{2}+r+{r}^{2}+\frac{1}{2}\left(-2r-1+1\right)-2$
${T} _{r}=3{r}^{2}+2r-1=\left(3r-1\right)\left(r+1\right)$
We have ${T} _{1}=\left(3-1\right)\left(1+1\right)=2.2$
${T} _{2}=\left(3\times2-1\right)\left(2+1\right)=5.3$
${T} _{3}=\left(3\times3-1\right)\left(3+1\right)=8.4$ which are in A.P
Thus,${T} _{n}=\left(3n-1\right)\left(n+1\right)$
From the above sequence, we note that
Product of even number and an odd number is Even
Product of odd number and an odd number is odd
Product of even number and an even number is Even
and we see that every term is a composite number.
Hence, their sum is a composite number.
$\therefore T$ is a composite number.

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The sum of all terms of the arithmetic progression having ten terms except for the first tens, is 99, and except for the sixth term, is 89. Find the third term of the progression if the sum of the first and the fifth term is equal to 10.

  1. 15

  2. 5

  3. 8

  4. 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:

${ S } _{ 10 }=99+{ T } _{ 1 }..........(i)\ { S } _{ 10 }=89+{ T } _{ 6 }..........(ii)$
where ${ S } _{ 10 }$ is the sum of $10$ terms of the A.P. and ${ T } _{ 1 }, { T } _{ 6 }$ are the first and sixth term respectively.
Say $a$ and $d$ are the first term and common difference of the A.P. respectively.
$\ \therefore { S } _{ 10 }=5\left{ 2a+9d \right} ;\quad { T } _{ 1 }=a;\quad { T } _{ 6 }=a+5d........(iii)\ \therefore 5\left{ 2a+9d \right} =a+99........(iv)\ 5\left{ 2a+9d \right} =a+89+5d........(v)\ $
Subtracting (iv) and (v), we get,
$10-5d=0\ =>d=2........(vi)$
Also given that
${ T } _{ 1 }+{ T } _{ 5 }=10\ =>a+a+4d=10\ =>2a+4\times 2=10\ =>2a=2\ =>a=1$
$\therefore { T } _{ 3 }=a+2d=1+2\times 2=5$

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

A progression of the form $a, ar, ar^2$, ..... is a

  1. geometric series

  2. harmonic series

  3. arithmetic progression

  4. geometric progression
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A progression of the form $a, ar, ar^2$, ..... is a geometric progression.
Geometric Progression refers to a sequence in which successor term of each term is obtained by multiplying a constant term.

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

Find the sum of an infinite G.P : $\displaystyle 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+.......$

  1. $\displaystyle \frac{3}{5}$
  2. $\displaystyle \frac{3}{2}$
  3. $\displaystyle \frac{49}{27}$
  4. $\displaystyle \frac{8}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given series is $1+ \dfrac {1}{3}+ \dfrac {1}{9 }+ \dfrac {1}{27}+......$

$a=1, r= \dfrac {1}{3}$

$\therefore S _{\infty}=\dfrac{a}{1-r}$

$S _{\infty}=\dfrac{1}{1-\dfrac{1}{3}}$

$S _{\infty}=\dfrac{1}{\dfrac{2}{3}}$

$\therefore S _{\infty}=\dfrac{3}{2}$

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

Find the GP whose $5^{th}$ term is $48$ and $9^{th}$ term is$ 768$.

  1. $3,6,12,24$
  2. $2,4,8,16$
  3. $6,12,24,48$
  4. $12,24,36,48$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle { ar }^{ 4 }=48$
$\displaystyle { ar }^{ 8 }=768$
$\displaystyle \therefore \quad { r }^{ 4 }=16$
$\displaystyle \therefore \quad r=2$
$\displaystyle a.{ 2 }^{ 4 }=48$
or, $\displaystyle a=\frac { 48 }{ 16 } =3$
The GP is 3,6, 12,24,.....

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

The reciprocals of all the terms of a geometric progression form a ________ progression.

  1. AP

  2. HP

  3. GP

  4. AGP
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let  $ a $ be the first term  and $ r $ be the common ratio of the GP. 

So, the series is $ a, ar, ar^2... $

Their reciprocals are $ \dfrac {1}{a}, \dfrac {1}{ar}, \dfrac {1}{ar^2} .. $

It is also a GP, with first term $ \dfrac {1}{a} $ and common ratio $ \dfrac {1}{r} $
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

In a _______ each term is found by multiplying the previous term by a constant.

  1. arithmetic sequence

  2. geometric series

  3. arithmetic series

  4. harmonic progression
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A geometric series is a series for which the ratio of each two consecutive terms is a constant function of the summation index .

Or,
In a Geometric series each term is found by multiplying the previous term by a constant.
$(Ans \to B)$

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

In a _______ each term is found by multiplying the previous term by a constant.

  1. geometric sequence

  2. arithmetic sequence

  3. geometric series

  4. harmonic sequence
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Sol:
We know that if a,b,c are in G.p then $b^2=ac$
We know that a G.P
$a,ar,ar^2.ar^3-----ar^n$
${ a } _{ 1 }{ ,a } _{ 2 },{ a } _{ 3 },{ a } _{ 4 },----{ a } _{ n }$
$\dfrac { { a } _{ 2 } }{ { a } _{ 1 } } =\dfrac { { a } _{ 3 } }{ { a } _{ 2 } } =\dfrac { { a } _{ 3 } }{ { a } _{ 3 } } ----\dfrac { { a } _{ n } }{ { a } _{ n-1 } } =r$  (r=constant)
Therefore in a geometric progression each term is found multiplying the previous term by constant .
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If a sequence of values follows a pattern of multiplying a fixed amount times each term to arrive at the following term, it is called a: 

  1. geometric sequence

  2. arithmetic sequence

  3. geometric series

  4. harmonic sequence
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$3,3^2,3^3,3^4,....(r=3)$
In a sequence if a fixed amount/constant is multiplied to each term to get the successive term the sequence is called geometric sequence.
Here $3$ is the constant which gets multiplied to each term to obtain the successive term.
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The number of terms in a sequence $6, 12, 24, ....1536$ represents a

  1. arithmetic progression

  2. harmonic progression

  3. geometric progression

  4. geometric series
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given series is $6,12,24,....1536$
Since, $\dfrac {12}{6} =2$ and $\dfrac {24}{12} =2$
i.e. the given sequence is a geometric sequence / progression. 
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

For which sequence below can we use the formula for the general term of a geometric sequence?

  1. $1, 3, 5, 7, 9.....$
  2. $2, 4, 6, 8, 10.....$
  3. $4, 8, 16, 32, 64....$
  4. $1, -1, 3, -2, 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a G.P., the ratio must be common throughout.
We use the formula for the general term of a geometric sequence for $4, 8, 16, 32, 64.... $
Here the common ratio is $2$.