Questions on Sequence and Series

Aptitude and ability tests are designed to assess your logical reasoning or thinking performance in a time bound manner.  It mainly consists of objective type questions.

Here are some questions on sequence and series for practice.

Directions for question number 1 to 10: In the given series, there is one term which does not belong to the given series. Find the term which does not belong to the given series.

  111      248      369      41664     525125

a. 111
b. 248
c. 369
d. 41664

Solution

111 = 1¹ 1² 1³; 248 = 2¹ 2² 2³; 41664 = 3¹ 3² 3³;         525125 = 5¹ 5² 5³

The 3^rd term is not in the sequence.

Hence (c)

2.    789      101112       131415     151617     181920

a. 101112
b. 151617
c. 181920
d. 131415

Solution

The numbers are three consecutive numbers. But the fourth term contains the no already used in the previous term. Hence it is not in the series.

Hence (b) 

3.      0          1          10        11        100      111

a.  111
b. 100
c. 0
d. 11

Solution

The numbers are consecutive terms of binary numbers. except the last term. It should be replaced by 101.

Hence (a)

4.      2          10        28        65        126      217

a. 2
b. 10
c. 65
d. 217

Solution

The numbers are cubes of the no.  + 1.

2 = 1³ + 1;        28 = 3³ + 1       65 = 4³ + 1

The second term does not fit in this series.

Hence (b)

5.     1427    2964    31625      425216     536343

a. 1427
b. 425216
c. 2964
d. 31625

Solution

The numbers are in the form a¹b²c³ starting from 1. The third term is not fit in the sequence. It should be replaced by 316125

Hence (d)

6.      12        20        30        36        46

a. 12
b. 20
c. 36
d. 30 

Solution

The difference between two consecutive numbers is increasing in a fashion of 2(+8, +10, +12, +14) except the 4^th term.

Hence (c)

7.      10        50        100      150      200

a. 10
b. 50
c. 150
d. 200

Solution

The common difference is 50 in all except the first term.

Hence (a)

8.      22        34        48        60        82

a. 22
b. 34
c. 60
d. 70 

Solution

The first digit is increasing by 1 and second digit is multiplied by 2. The 4^th term is not satisfying the given condition. So in the place of 60 the number is 66 (for first place 4+1=5 and for last place 8*2 = 16 which gives 1 as carry to 5 n makes it 6)

Hence (c)

9.      135      7911     131415      171921     232527

a. 135
b. 7911
c. 131415
d. 171921

Solution

The numbers are three consecutive odd numbers. The third term has an even no in between.

Hence (d)

10    147      101316      172023     252729     313437

a. 147
b. 172023
c. 252729
d. 101316

Solution

Each no has three parts (difference of three is there). The 2^nd part and 1^st part in all is 3 and that of 3^rd and 2^nd part is also 3.

But in 252729 the difference is 2.

Hence (c)