Mathematics · Quantitative Aptitude

Number Sums and Series

268 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 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.
$S=\dfrac{a-ar^{n+1}}{1-r}$
$S=\dfrac{4-\frac{1}{32}}{1-\frac{1}{2}}$
$S=\dfrac{\frac{127}{32}}{\frac{1}{2}}$
$S=\dfrac{127}{16}$

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

The sum of the squares of three numbers which are in the ratio $2 : 3 : 4$ is $725.$ What are these numbers?

  1. $10, 15, 20$
  2. $14, 21, 28$
  3. $20, 15, 30$
  4. $20, 30, 40$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the three numbers be 2x, 3x and 4x.
Given, $(2x)^2 + (3x)^2 + (4x)^2 = 725$
$\Rightarrow 4x^2 + 9x^2 + 16 x^2 = 725 \Rightarrow 29 x^2 = 725$
$\Rightarrow x^2 = 25 \Rightarrow x = 5$
$\therefore$ The numbers are 10, 15 and 20.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

The whole number whose sum is $72$ cannot be in the ratio

  1. $\dfrac{5}{7}$
  2. $\dfrac{3}{5}$
  3. $\dfrac{3}{4}$
  4. $\dfrac{4}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The whole number must   be divided by sum of two ratio particularly

A.$5+7=12$
$72$ can be divided by $12$

B.$ 3+5=8$
$72$ can be divided by $8$

C.$ 4+3=7$
$72$ can not be divided by $7$

D.$4+5=9$
$72$ can be divided by $9$