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 average arithmetic mean of ap introduction to averages means

The  Sum of three numbers in AP is $75$, and product of extremities is $609$. The numbrs and AM of 1st two numbers is 

  1. $\{21,25,29\}$, AM $= 23$
  2. $\{13,17,21\}$, AM $= 22$
  3. $\{21,25,29\}$, AM $= 25$
  4. $\{21,22,29\}$, AM $= 23$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the numbers in A.P.  be (a-d), a, (a+d)

$\therefore (a-d)+a+(a+d)=75$
$\Rightarrow a=25$
Also, $(a-d)(a+d)=609$
$\Rightarrow a^{2}-d^{2}=609$
$\Rightarrow 25^{2}-d^{2}=609$
$\therefore d=\pm4$
The numbers are {21,25,29} or {29,25,21}
According to option, we take {21,25,29}
A.M. of 1st two numbers $=\dfrac{21+25}{2}=23$

Multiple choice maths average arithmetic mean of ap introduction to averages means

The sum of four numbers in AP is $176$. The product of 1st and last is $1855$.  The mean of middle two is 

  1. $42$
  2. $41$
  3. $44$
  4. $53$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the numbers in A.P. be  (a-3d), (a-d), (a+d), (a+3d)
$\therefore (a-3d)+(a-d)+(a+d)+(a+3d)=176$
$\Rightarrow 4a=176$
$\therefore a=44$
$\therefore$ The mean of middle two is:
$=\dfrac{(a-d)+(a+d)}{2}$
$=a=44$                           
Multiple choice maths fraction lowest form of a fraction simplest ratio lowest form of fractions

The number $2.525252$ can be written as a fraction, when reduced to the lowest term, the sum of the numerator and denominator is:

  1. $7$
  2. $29$
  3. $141$
  4. $349$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the given number be $x=2.525252....$
multiplying with $100$ on both sides
$\Rightarrow 100x=252.525252...$
$\Rightarrow 100x=250+2.5252...$
$\Rightarrow 100x=250+x\Rightarrow 99x=250$
$\Rightarrow x=\dfrac{250}{99}$
$\therefore$ Sum of numerator and denominator $=25099=349$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The sum of the three numbers in A.P is $21$ and the product of the first and third number of the sequence is $45$. What are the three numbers?

  1. $5, 7$ and $9$
  2. $9, 7$, and $5$
  3. $3, 7$, and $11$

  4. Both (1) and (2)

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the numbers are be $a - d, a, a + d$
Then $a - d + a + a + d = 21$
$3a = 21$
$a = 7$
and $(a - d)(a + d) = 45$
$a^2 - d^2 = 45$
$d^2 = 4$
$d=\pm 2$
Hence, the numbers are $5, 7$ and $9$ when $d = 2$ and $9, 7$ and $5$ when $d = -2$. In both the cases numbers are the same.

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

The sum of $10$ numbers is $100$. The first term is $1$. Find its common difference.

  1. $2$
  2. $1$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The sum of first $n$ terms of arithmetic series formula can be written as,
$S _n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)
$n =$ number of terms $= 10$
$S _n = 100$
First term, $a = 1$
Common difference, $d = ?$
From $(1)$, we have
$100 = \dfrac{10}{2} [2 \times 1 + (10 - 1)d]$
$100 =  5[2 + 9d]$
$100 = 10 + 45d$
$ 100 - 10 = 45d$
$90 = 45d$
$d = \dfrac{90}{45} = 2$
The common difference is $2$.
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The sum of all the natural numbers from $200$ to $600$(both inclusive) which are neither divisible by $8$ nor by $12$ is?

  1. $123968$
  2. $133068$
  3. $133268$
  4. $187332$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$D(8)=$ numbers divisible by $8 = 200, 208, 216, 224, 232, 240,.., 592, 600.$
Total $D(8)$ numbers $= 51$
Sum of $D(8)$ numbers $=\left[\dfrac{51}{2}\times (200 + 600)\right] = (51\times 400) = 20400$
 
$D(12)=$ numbers divisible by $12 = 204, 216, 228, 240, 252, 264,.., 588, 600.$
Total $D(12)$ numbers $=34$
Sum of $D(12)$ numbers $=\left[\dfrac{34}{2}\times (204 + 600)\right] = (17\times 804) = 13668$

Now, $D(8\cap 12) =$ numbers divisible by both $8$ and $12 = 216, 240, 264,..., 576, 600.$

Total $D(8\cap 12)$ numbers $=17$

Sum of $D(8$ intersect $12)$ numbers $= \left[\dfrac{17}{2}\times (216 + 600)\right] = (17\times 408) = 6936$

So, $D(8\cup 12) =$ numbers divisible by either $8$ or $12$

                         $= D(8) + D(12) - D(8\cap 12).$

So, sum of $D(8\cup 12)$ numbers $= 20400 + 13668 - 6936 = 27132.$

Now, sum of all natural numbers ranging from $200$ to $600 = \left[\dfrac{401}{2}\times (200 + 600)\right] = (401\times 400) = 160400.$

Sum of all natural numbers from $200$ to $600$ which are neither divisible by $8 $nor by $12 = (160400 - 27132)$
                                                                                                                                                $ = 133268.$
Multiple choice

What is the sum of the first 100 positive integers?

  1. 4950

  2. 5050

  3. 5150

  4. 5250
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The sum of the first 100 positive integers can be found using the formula (S = \frac{n(n+1)}{2}), where n is the number of integers. Substituting n = 100 into the formula, we get: (S = \frac{100(100+1)}{2} = \frac{100 \cdot 101}{2} = 5050). Therefore, the sum of the first 100 positive integers is 5050.

Multiple choice

What is the sum of the first 100 positive integers?

  1. 5050

  2. 5150

  3. 5250

  4. 5350
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the first n positive integers is given by the formula n(n+1)/2. Substituting n = 100, we get 100(101)/2 = 5050.

Multiple choice

What is the sum of the first 50 positive even integers?

  1. 2550

  2. 2650

  3. 2750

  4. 2850
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the first 50 positive even integers is 2550.

Multiple choice

What is the sum of the first 100 positive integers?

  1. 5050

  2. 5150

  3. 5250

  4. 5350
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the first n positive integers is given by the formula n(n+1)/2. Therefore, the sum of the first 100 positive integers is 100(101)/2 = 5050.

Multiple choice

In a certain country, the sum of the first 100 positive integers is equal to the sum of the first 100 even integers. What is the sum of the first 100 odd integers?

  1. 10000

  2. 10100

  3. 10200

  4. 10300
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The sum of the first 100 positive integers is (1 + 2 + 3 + ... + 100) = (100 * 101) / 2 = 5050. The sum of the first 100 even integers is (2 + 4 + 6 + ... + 200) = (100 * 202) / 2 = 10100. Therefore, the sum of the first 100 odd integers is 10100 - 5050 = 5050.