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 even and odd numbers sum of numbers odd and even numbers

Sum of two even integers is :

  1. Even

  2. Odd

  3. Both

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
A 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:


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

Now let us take two even integers say, $2$ and $4$, then their sum will be $2+4=6$ which is also an even integer.

Hence, the sum of even integers is always even.

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

Find three consecutive even integers such that the sum of first two integers is same as the sum of third integer and $6$.

  1. $4,6,8$
  2. $6,8,10$
  3. $8,10,12$
  4. $10,12,14$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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


Now, it is given that the sum of first two integers is same as the sum of the third integer and $6$ which means:

$x+(x+2)=(x+4)+6\ \Rightarrow 2x+2=x+10\ \Rightarrow 2x-x=10-2\ \Rightarrow x=8$

Therefore, the first even integer is $8$ then the second integer is $x+2=8+2=10$ and the third integer is $x+4=8+4=12$

Hence, the three consecutive even integers are $8,10,12$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

The two consecutive multiples of $3$ whose sum is $51$ are __________.

  1. $24, 27$
  2. $20, 31$
  3. $40, 11$
  4. $25, 26$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lets say $x$ is the multiple of $3$

Next consecutive multiple of $3$ will be $(x+3)$
Given sum is $=51$
$\Rightarrow x+x+3=51$
$\Rightarrow  2x=48$
$\Rightarrow x=24$
Two consecutive multiples of $3$ are $24,27$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

If the sum of four consecutive even integers is $212$, what is the value of the second even integer?

  1. $50$
  2. $51$
  3. $52$
  4. $53$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the four consecutive even numbers be $x, x + 2, x + 4$ and $x + 6$.

Therefore, $x + x + 2 + x + 4 + x + 6 = 212$
$4x + 12 = 212$
$4x = 212 - 12$
$4x = 200$ (Divide both sides by $4$)
$x = 50$
The second number be $x + 2$
So, the second number is $52$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

If the sum of four consecutive odd integers is $400$, what is the value of the first odd integer?

  1. $95$
  2. $96$
  3. $97$
  4. $98$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the four consecutive odd numbers be $x, x + 2, x + 4$ and $x + 6$.
Therefore, $x + x + 2 + x + 4 + x + 6 = 400$
$4x + 12 = 400$
$4x = 400 - 12$
$4x = 388$  (Divide both sides by $4$)
$x = 97$
The first number be $x$ 
So, the first number is $97$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

If the sum of four consecutive integers is $110$, what is the value of the third consecutive integer?

  1. $26$
  2. $27$
  3. $28$
  4. $29$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the four consecutive numbers be $x, x + 1, x + 2$ and $x + 3$.
Therefore, $x + x + 1 + x + 2 + x + 3 = 110$
$4x + 6 = 110$
$4x = 110 - 6$
$4x = 104$   (Divide both sides by $4$)
$x = 26$
The third number be $x + 2$ 
So, the third number is $28$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

The sum of two numbers is $45$ and their difference is $11$. What are the two numbers?

  1. $28$ and $17$
  2. $27$ and $18$
  3. $25$ and $20$
  4. $22$ and $23$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the numbers be $a$ and $b$
Given that 

$a+b=45$ ....(1)
$a-b=11$ ....(2)
Adding these two equations, we get
$2a=56$
$\Rightarrow a=28$
Substituting value of $a$ in equation (1), we get
$28+b=45$
$\Rightarrow b=45-28$
$\Rightarrow b=17$
We get $a = 28$ and $b=17$

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

$\sum{n^3}=$

  1. $(\sum{n})^3$
  2. $(\sum{n})^2$
  3. $(\sum{n})^3+(\sum{n})^2$
  4. $\sum{(n+n^2)}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have, $ \sum { { n }^{ 3 } } = { 1 }^{ 3 }+{ 2 }^{ 3 }+....+{ n }^{ 3 } $
The formula to find the sum of cubes of natural numbers is $ ={( \dfrac {n(n+1)}{2} )}^2 $

But sum of first $ n $ natural numbers is$  \sum { { n } } = \dfrac {n(n+1)}{2} $ 

So, $ \sum { { n }^{ 3 } } = { (\sum { { n } })}^2 $
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

The sum of two nonzero numbers is $8$. The minimum value of the sum of their reciprocals is

  1. $\displaystyle \frac{1}{4}$
  2. $\displaystyle \frac{1}{2}$
  3. $\displaystyle \frac{1}{8}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $x$ and $y$ be two numbers 
$\Rightarrow x+y = 8$
Assume $z$ be be sum of their inverse
$z = 1/x+1/y =\dfrac{x+y}{xy} = \dfrac{8}{xy} = \dfrac{8}{x(8-x)}$
For minimum value of $z $
$\dfrac{dz}{dx} = 0 =\dfrac{16(4-x)}{(x(8-x))^2}\Rightarrow x = 4$
Hence minimum value of $z$ is $=1/4+1/4 = \dfrac{1}{2}$ 

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Two parts of $64$ such that the sum of their cubes is minimum will be-

  1. $44, 20$
  2. $16, 48$
  3. $32, 32 $
  4. $50, 14$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let one part be $x$.
Hence another part be $(64-x)$
Thus 
Their cubes will be 
$x^{3}+(64-x)^{3}=y$
Thus 
$y'=3x^{2}-3(64-x)^{2}=0$
Or 
$x^{2}=(64-x)^{2}$
Or 
$x=64-x$ and $x=-64+x$
Hence
$x=32$.
Hence both the parts are 
$32,32$.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

The sum of two numbers is 6. The minimum value of the sum of their reciprocals is

  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{6}{5}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{2}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x+y=6$
$Sum =\dfrac {1}{x}+\cfrac {1}{y}$
$\dfrac {d(sum)}{dx}=\dfrac {-1}{x^2}+\dfrac {1}{(6-x)^2}=0$
$x^2=(6-x)^2$
$x=\pm (6-x)$
$x=3$,  $y=3$
$Sum =\dfrac {2}{3}$