Mathematics · Quantitative Aptitude

Number Sums and Series

287 Questions

Number sums and series questions test the ability to find patterns and calculate the sum of number sequences. These problems are a staple in the quantitative aptitude sections of various competitive exams. Practice this collection to improve speed and accuracy in solving arithmetic and geometric series problems.

Sum of consecutive integersGeometric series sumsRatio and proportion sumsOdd and even number propertiesRoman numeral calculations

Number Sums and Series Questions

Multiple choice maths negative numbers and integers odd and even numbers operations on integers different types of numbers

If the number of consecutive odd integers whose sum can be expressed as $50^2 - 13^2$ is k then k, can be 

  1. 33

  2. 35

  3. 37

  4. 39
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sum of odd $n$ consecutive numbers $n^2$

$\therefore (1+3+5\dots\dots (2n-1))=n^2$
where $n$ represents the number of terms.
$\therefore 50^2=1+3+5\dots 99=50\text{ }terms$
$\therefore 13^2=1+3+5\dots 25=13\text{ }terms$
$\therefore 50^2-13^2$$=(1+3+5\dots 99)-(1+3+5\dots 25)\=(27+29\dots 99)\ =37\text{ }terms.$

Multiple choice maths negative numbers and integers odd and even numbers operations on integers different types of numbers

The sum of even numbers between $1$ and $31$ is:

  1. $6$
  2. $28$
  3. $240$
  4. $512$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let ${S} _{n}=(2+4+6+.....+30)$. This is an A.P in which $a=2,d=2$ and $l=30$
Let the number of terms be $n$. Then,
$a+(n-1)d=30$
$\Rightarrow$ $2+(n-1)\times 2=30$
$\Rightarrow$ $n=15$
$\therefore$ ${S} _{n}=\cfrac{n}{2}(a+l)=\cfrac{15}{2}\times (2+30)=(15\times 16)=240$.

Multiple choice maths hcf-lcm introduction to multiples multiples lcm

The sum of the first five multiples of $6$  is

  1. $90$
  2. $60$
  3. $30$
  4. $120$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

first five multiple of 6 are 

$6\times 1=6$
$6\times 2=12$
$6\times 3=18$
$6\times 4=24$
$6\times 5=30$
Their sum will be $6+12+18+24+30=90$
So correct answer will be option A

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

If the sum of the first n integers is 15. What is n? (use Gauss method)

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Gauss formula,
$S _n = \dfrac{n}{2} (n + 1)$
$15 = \dfrac{n}{2} (n + 1)$
$30 = n (n + 1)$
$30 = n^2 + n$   ....... (1)
By factorization we can write equation (1) as
$n^2 - 6n + 5n - 30 = 0$
$n(n - 6) + 5 (n - 6) = 0$
$n = 6, - 5$
$\therefore$ The n integers will start from -5 to 6.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If the sum of two integers is $-2$ and their product is $-24$, the numbers are

  1. $6$ and $4$
  2. $-6$ and $4$
  3. $-6$ and $-4$
  4. $6$ and $-4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $p$ and $q$ be the required integers

If $p+q$ and $pq$ are known then quadratic equation corresponding to roots as $p$ and $q$ is given by,
$x^2-(p+q)x+pq=0$
$\Rightarrow x^2+2x-24=0$, substitute the given values
$\Rightarrow x^2+6x-4x-24=0$, split the middle term
$\Rightarrow (x^2+6x)+(-4x-24)=0$, group pair of terms
$\Rightarrow x(x+6)-4(x+6)=0$, factor each binomials 
$\Rightarrow (x+6)(x-4)=0$, factor out common factor 
$\Rightarrow x=-6$ or $x=4$, set each factor to $0$

Hence $p$ and $q$ are $-6$ and $4$
Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The sum of first 10 natural numbers is____.

  1. 100

  2. 55

  3. 50

  4. 90
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The series of first 10 natural numbers is an arithmetic progressions with first tern as 1 and common difference 1. So the sum of the series will be Sn = n/2 { 2a+ ( n-1 ) d } where n is the number of terms in the series, a is the first term and d is the common difference.

S10= 10/2 { 2(1) + ( 10-1 ) 1 }

     = 5 ( 2+9)

    = 5 ( 11 )

    = 55

Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

The sum of two numbers is $4000$. $10\%$ of one number is $\displaystyle 6\frac{2}{3}$ $\%$ of the other The difference of the number is

  1. $600$
  2. $800$
  3. $1025$
  4. $1175$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let one number be $x$. 

Then the other number $= 4000 - x$
Given, $10\%$ of $\displaystyle x=6\frac{2}{3}\%$ of $ (4000-x)$
$\displaystyle \Rightarrow \frac{10}{100}\times x=\frac{20}{3}\times \frac{1}{100}\times (4000-x)$
$\Rightarrow 10x=\dfrac{20}{3}\times 4000-\dfrac{20x}{3}$
$\displaystyle \Rightarrow 10x+\frac{20x}{3}=\frac{20}{3}\times 4000$
$\Rightarrow \dfrac{50x}{3}=\dfrac{20}{3}\times 4000$
$\displaystyle \Rightarrow x=\frac{20\times 4000}{50}=1600$
The two numbers are $1600$ and $2400$. 
$\displaystyle \therefore$ Their difference is $2400 - 1600 = 800$.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Given that the sum of the odd integers from $1$ to $99$ inclusive is $2500$, what is the sum of the even integers from $2$ to $100$ inclusive?

  1. 2450

  2. 2550

  3. 2460

  4. 22500
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Odd integers $=1,3,5,7,9.....$

Sum of odd integers $=2500$
Even integers $=2,4,6,8,,10......$
In the series of even integers each term is one more than the each term of odd integers.
Hence, there are $50$ terms.
So, the sum of even integers $=$ $2500+50=2550$

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Find three consecutive odd integers such that the sum of first and third integers is same as the second integer when decreased by $9$.

  1. $-9,-7,-5$
  2. $-13,-11,-9$
  3. $-15,-13,-11$
  4. $-11,-9,-7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let us say the first odd integer be $x$. The second consecutive odd integer would be $x+2$ (zit would not be $x+1$ because that would result in an even integer. The sum of two odd integers is even). The third consecutive odd integer would be $(x+2)+2$ or $x+4$.


Now, it is given that the sum of first and third integers is same as the second integer when decreased by $9$ which means:

$x+(x+4)=(x+2)-9\ \Rightarrow 2x+4=x-7\ \Rightarrow 2x-x=-7-4\ \Rightarrow x=-11$

Therefore, the first odd integer is $-11$ then the second integer is $x+2=-11+2=-9$ and the third integer is $x+4=-11+4=-7$

Hence, the three consecutive odd integers are $-11,-9,-7$.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Sum of one odd and one even integers is :

  1. Even

  2. Odd

  3. Both

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

An even integer is an integer that is evenly divisible by $2$, that is, division by $2$ results in an integer without any remainder. The set of even integers is:


$.....-8,-6,-4,-2,2,4,6,8,.....$

Whereas, an odd integer is an integer that is not divisible by $2$. The set of odd integers is:

$.....-9,-7,-5,-3,-1,1,3,5,7,9.....$

Now let us take an even integer say, $2$ and an odd integer say $3$, then their sum will be $2+3=5$ which is an odd integer.

Hence, the sum of one odd and one even integer is always odd.