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 average arithmetic mean of ap introduction to averages means

The mean  of five numbers in AP is $89$. The product of first and last terms is $7885$. The AM of first, third and fifth term is

  1. $83$
  2. $86$
  3. $89$
  4. $90$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the numbers in A.P. be  (a-2d), (a-d), a, (a+d), (a+2d)
$\therefore \dfrac{(a-2d)+(a-d)+a+(a+d)+(a+2d)}{5}=89$
$\Rightarrow \dfrac{5a}{5}=89$
$\therefore a=89$
$\therefore$ The AM of 1st, 3rd & 5th term is:
$=\dfrac{(a-2d)+a+(a+2d)}{3}$
$=a=89$                           
Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean of 1, 8, 27, 64, ....... up to n terms is given by

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

Solution:

Given:
$1,8,27,64,...........$ upto $n$ terms
or,$1,2^3,3^3,4^3,..............$ upto $n$ terms
$\therefore AM=\cfrac{1+2^3+3^3+4^3............+n^3}{n}=\cfrac{\left[\cfrac{n(n+1)}2\right]^2}{n}$
$=\cfrac{n^2(n+1)^2}{4n}=\cfrac{n(n+1)^2}4$
Hence, C is the correct option.

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

The $n^{th}$ term of the series $3+7+14+24+.....$ is

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

The sequence is 3, 7, 14, 24. The differences are 4, 7, 10. This is a quadratic sequence. Testing n=1 gives 3, n=2 gives 7. Option B fits the pattern.

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

If $T _n$ denotes the nth terms of the series.$ 2+3+6+11+18+..........$ , then $T _50$ is :

  1. $49^2-1$
  2. $49^2$
  3. $50^2+1$
  4. $49^2+2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$2+3+6+11+18+.................$

$\begin{matrix} { T _{ n } }=a{ x^{ 2 } }+bx+c \ { T _{ 1 } }=a+b+c=2 \ { T _{ 2 } }=a+b+c=3 \ { T _{ 3 } }=a+b+c=6 \ 3a+b=1 \ 5a+b=3 \ 2a=2 \ a=1 \ b=-2 \ c=3 \  \end{matrix}$
${T _n} = {n^2} - 2n + 3$
${T _{50}} = {\left( {50} \right)^2} - 2\left( {50} \right) + 3$
${T _{50}} = {\left( {49} \right)^2} + 2$
Hence,
Option $D$ is correct answer.

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

In an arithmetic progression the sum of two terms equidistant from the beginning and the end is always _____ to the sum of the first and last terms.

  1. equal

  2. unequal

  3. different

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

Let $a$ be first term $d$ be common difference of an AP having $n$ number of terms.

So $m$th term from the beginning is, $a _m=a+(m-1)d$
and $m$ the term from end is $a _{n-m}= a+(n-1)d-(m-1)d=a+(n-m)d$
So $a _m+a _{n-m}=2a+(n-1)d=[a]+[a+(n-1)d]=$ sum of first and last term 

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

In which property the sum of two terms equidistant from the beginning and the end is always same or equal to the sum of the first and last terms?

  1. A.P.

  2. G.P.

  3. H.P.

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

Let $a$ be first term $d$ be common difference of an AP having $n$ number of terms.

So $m$th term from the beginning is, $a _m=a+(m-1)d$
and $m$ the term from end is $a _{n-m}= a+(n-1)d-(m-1)d=a+(n-m)d$
So $a _m+a _{n-m}=2a+(n-1)d=[a]+[a+(n-1)d]=$ sum of first and last term 

Hence option 'A' is correct choice 

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

The sum of first $10$ terms and $20$ terms of an AP are $120$ and $440$ respectively. What is the first term?

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$\implies 120 =5(2a+9d)$
$\implies 24=2a+9d$ .............. $(i)$
Sum of first 20 terms $=\frac { 20 }{ 2 } (2a+(20-1)d)$
$\implies 440 =10(2a+19d)$
$\implies 44=2a+19d$ ......... $(ii)$
Subtracting equation (i) from (ii) gives
$20=10d$
$\implies d=2$
Common difference =2
Substituting in $(i)$, we get 
$a=3$
Hence, option B is correct

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

Which term of A.P. $20, 19\displaystyle\frac{1}{4}, 18\frac{1}{2}$,..... is first negative term?

  1. $!8$th
  2. $15$th
  3. $28$th
  4. $27$th
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$20, \displaystyle 19\frac{1}{4}, 18\frac{1}{2}, ....$
or $\displaystyle 20, \frac{77}{4}, \frac{37}{2}, ....$
$a=20$
$d=\displaystyle\frac{77}{4}-20=\frac{-3}{4}$
Let $n^{th}$ term of A.P. be first negative term
So, $20+(n-1)\left(\displaystyle\frac{-3}{4}\right)<0$
$\Rightarrow 80-3n+3<0$
$\Rightarrow 3n>83$
$\Rightarrow n > 27\displaystyle\frac{2}{3}$
Hence, $28^{th}$ term is first negative term.
(Option $3$).

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

An AP consists of $15$ terms. The three middle most terms is $69$ and the last three terms is $123$. Find the A.P.

  1. 2,5,8,11.....44

  2. 2,4,8,....62

  3. 2,3,4....16

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The $three$ middle most terms of $15$ terms is $69.$

${a} _{7}+{a} _{8}+{a} _{9}=69$
${a} _{13}+{a} _{14}+{a} _{15}=123$
$a+6d+a+7d+a+8d=69\Longrightarrow 3a+21d=69$
$a+12d+a+13d+a+14d=123\Longrightarrow 3a+39d=123$

$3a+21d=69$
$3a+39d=123$

          $18d=54,\d=3,\a=2.$
$2,5,8,11,\dots 44.$  

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

The largest term to common to the sequences $1,11,21,31,... to 100$ terms and $31,36,41,46, ...to 100$ terms is 

  1. $511$
  2. $471$
  3. $281$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 The first sequence is $1+11+21+31+41+51+61+71+81+91+...............$ upto $100th$ terms
last term will be =$1+(100-1)\times 10=991$
and for the second sequence  $31+36+41+46+51+56+.................$ upto $100th$ term 
last term will be=$31+(100-1)\times 5=526$
Now,
we have the common terms as $31+51+71$ and $n$ term of this sequence wil be =$31+(n-1)\times 20$
$=>20n+11$
and this will be less than $526$
So,
$=>20n+11<526$
$=>20n<526-11$
$=>n<\dfrac{515}{20}=25.75$
so $n=25$
Hence the largest common term is =$20\times 25\times 11=511$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $9^{th}$ term of an A.P. be zero then the ratio of its $2022^{th}$ and $10^{th}$ term is....... 

  1. $2013:1$
  2. $1:2013$
  3. $2013:8$
  4. $8:2013$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
General term of $A.P\ \Rightarrow \ at (n-1)d$
$\therefore \ a+(9-1)d=0$
$9+8d=0$
$a=-8d$
$\therefore \ \dfrac {a+(2022-1)d}{a+(10-1)d}$
$\Rightarrow \ \dfrac {-8d+2021d}{-8d+9d}$
$=\dfrac {2013d}{d}=\boxed {2013:1}$
$2013:1$, Option $A$ is correct

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

If the ratio of sum of n terms of two sequences is (3n+8):(7n+15), then the ratio of their $ 12^{th}$ term is ---------.

  1. 16 : 7

  2. 7 : 16

  3. 74 : 169

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

Let $a,A$ be first term and $d,D$ be common difference and $S _n,s _n$ be there respective sum.

$\cfrac{s _n}{S _n}=\cfrac{(n/2)[2a+(n-1)d]}{(N/2)[2A+(N-1)D]}=\cfrac{3n+8}{7n+15}$
Put $n=23$, we get
$\cfrac{a+11d}{A+11D}=\cfrac{77}{176}=\cfrac{t _{12}}{T _{12}}$

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

Let ${a _1},{a _2},{a _3}.....$ and ${b _1},{b _2},{b _3}......$ be AP such that ${a _1}=25,{b _1}=75$ and ${a _{100}} + {b _{100}} = 100$. Then,

  1. The difference between successive terms in progression $a$ is opposite the difference in progression $b$
  2. ${a _n} + {b _n} = 100$ for any n
  3. $({a _1} + {b _1}),({a _2} + {b _2}),({a _3} + {b _3}),....$ are in AP
  4. $\sum\limits _{r = 1}^{100} {({a _r} + {b _r})} = 10000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a_n and b_n are APs, their sum a_n + b_n is also an AP. The common difference of the sum is the sum of the individual common differences.

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

The sum of first $10$ terms and $20$ terms of an AP are $120$ and $440$ respectively. What is the common difference?

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

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$\implies 120 =5(2a+9d)$
$\implies 24=2a+9d$ .............. $(i)$
Sum of first 20 terms $=\frac { 20 }{ 2 } (2a+(20-1)d)$
$\implies 440 =10(2a+19d)$
$\implies 44=2a+19d$ ......... $(ii)$
Subtracting equation (i) from (ii) gives
$20=10d$
$\implies d=2$
Common difference =2
Hence, option B is correct.