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 fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

$96^{2}-1$ is a product of two consecutive odd numbers. Find those two odd numbers.

  1. 96 and 98

  2. 93 and 95

  3. 95 and 97

  4. 99 and 101
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$96^{2}-1 = 9215$
we can express the above number into general form $a^{2}-1 = (a + 1)\times (a - 1)$
Where a = 96
So, $96^{2}-1 = (96 + 1)\times (96 - 1)$
= $97 \times 95 = 9215$
Therefore, the two odd consecutive numbers are 95 and 97.

Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

Find the series and also find the total of first 10 consecutive odd numbers.

  1. $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100$
  2. $1 + 3 + 5 + 7 + 9 + 11 + 13 + 16 + 17 + 19 = 101$
  3. $1 + 3 + 5 + 7 + 10 + 11 + 13 + 15 + 17 + 19 = 101$
  4. $1 + 3 + 6 + 6 + 9 + 11 + 13 + 15 + 17 + 19 = 100$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of first 10 consecutive odd numbers $= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100$
Since, we know the formula for sum of consecutive odd numbers = $n^2$
So, $n = 10$, Sum $= 10^2 = 100$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Find the sum of the infinite geometric series $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.......$

  1. $16$
  2. $14$
  3. $-11$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given sequence is $1+\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}+....$
Thus $a=1$, $r = \dfrac{1}{2}$
Therefore, $\text{sum} =\dfrac{a}{1-r}$
$\Rightarrow \text{sum} = \dfrac{1}{1-\frac{1}{2}}$
$\Rightarrow \text{sum} =2$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The sum of  $3,1,\dfrac 13 ,....$ is

  1. $\dfrac 52$
  2. $\dfrac 92$
  3. $\dfrac 72$
  4. $\dfrac {11}2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given series is $3,1,\dfrac 13,..$


Given series is in GP.

The common ratio is given as $\dfrac{1}{3}$

The sum of infinite terms is  $\dfrac{a}{1-r}$

$\implies  \dfrac 3{1-\dfrac 13}$

$\implies \dfrac{3}{\dfrac 23}=\dfrac 92$


Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Three numbers whose sum is $45$ are in A.P. If $5$ is subtracted from the first number and $25$ is added to third number, the numbers are in G.P. Then numbers can be

  1. $10,\ 15,\ 20$
  2. $8,\ 15,\ 22$
  3. $5,\ 15,\ 25$
  4. $12,\ 15,\ 18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the numbers be $a-d,a,a+d$

Their sum is $45\a-d+d+a+d=45\3a=45\a=15$
The changed numbers are $15-d-5,15,15+d+25\10-d,15,40+d$
Condition to be in GP is $b^2=ac\15^2=(10-d)(40+d)\225=400-30d-d^2\d^2+30d-175=0\d^2+35d-5d-175=0\d(d+35)-5(d+35)=0\(d-5)(d+35)=0\d=5,-35$
$a=15,d=5$
So the series is $10,15,20$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Find the sum of the geometric series $4 + 2 + 1 +... +$ $\dfrac{1}{16}$

  1. $\dfrac{17}{16}$
  2. $\dfrac{107}{16}$
  3. $\dfrac{117}{16}$
  4. $\dfrac{127}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given series is $4+2+1+.....+\dfrac {1}{16}$
First term, $a$ is $4$
Common ratio, $r =$ $\dfrac{2}{4}=\dfrac{1}{2}$
Use the formula for the sum of the geometric series.
$ar^n$ is a next term.
$\dfrac{1}{16}=\dfrac{1}{2}\times \dfrac{1}{16}=\dfrac{1}{32}$ is the next term.
$S=\dfrac{a-ar^{n+1}}{1-r}$
$S=\dfrac{4-\frac{1}{32}}{1-\frac{1}{2}}$
$S=\dfrac{\frac{127}{32}}{\frac{1}{2}}$
$S=\dfrac{127}{16}$