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Sequence and Series – Definition, Formulas | Problems and Solutions with Tricks

Sequences and Series

Group of numbers arranged in order by some fixed rule is called as sequences. When a sequence follows a particular rule, it is called progress. It is not the same as a series that is defined as the sum of the elements of a sequence.

Read the article to know about Sequences and Series Definition from this page. Students who are preparing for competitive examination can check Sequences and Series Formulas here. You can download Sequences and Series Problems and Solutions with Tricks from below section of this page.

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Sequences and Series

Sequences and Series Definition

Sequences:

 A set of numbers arranged in order by some fixed rule is called as sequences.

 For example

 (i) 2, 4, 6, 8, 10, 12, 14 – – – – – – – – – – – –

 (ii) 1, 3, 5, 7, 9, – – – – – – – – – – – –

In sequence a1, a2, a3, …… an, ….. a1 is the first term, a2 is the second term, a3 is the third and so on.

A sequence is called finite sequence if it has finite terms e.g., 2, 4, 6, 8, 10, 12, 14, 16.

A sequence is called infinite sequence if it has infinite terms, e.g. 4, 6, 8, 10, 12, 14, – – – – – – – –

Series:

A series can be highly generalized as the sum of all the terms in a sequence. However, there has to be a definite relationship between all the terms of the sequence.

Types of Sequence and Series

Some of the most common examples of sequences are:

  • Arithmetic Sequences
  • Geometric Sequences
  • Harmonic Sequences
  • Fibonacci Numbers

Arithmetic Sequences

A sequence in which every term is created by adding or subtracting a definite number to the preceding number is an arithmetic sequence.

Example:

1, 4, 7, 10, 13, 16, 19, 22, 25, …

This sequence has a difference of 3 between each number.

Its Rule is xn = 3n-2

Geometric Sequences

A sequence in which every term is obtained by multiplying or dividing a definite number with the preceding number is known as a geometric sequence.

Example:

2, 4, 8, 16, 32, 64, 128, 256, …

This sequence has a factor of 2 between each number.

Its Rule is xn = 2n

Harmonic Sequences

A series of numbers is said to be in harmonic sequence if the reciprocals of all the elements of the sequence form an arithmetic sequence.

Example:

1, 4, 9, 16, 25, 36, 49, 64, 81, …

The next number is made by squaring where it is in the pattern.

Rule is xn = n2

Fibonacci Numbers

Fibonacci numbers form an interesting sequence of numbers in which each element is obtained by adding two preceding elements and the sequence starts with 0 and 1. Sequence is defined as, F0 = 0 and F1 = 1 and Fn = Fn-1 + Fn-2.

Example:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

Rule is xn = xn-1 + xn-2

Sequence and Series Formulas

List of some basic formula of arithmetic progression and geometric progression are

Arithmetic ProgressionGeometric Progression
Sequencea, a+d, a+2d,……,a+(n-1)d,….a, ar, ar2,….,ar(n-1),…
Common Difference or RatioSuccessive term – Preceding term

Common difference = d = a2 – a1

Successive term/Preceding term

Common ratio = r = ar(n-1)/ar(n-2)

General Term (nth Term)an = a + (n-1)dan = ar(n-1)
nth term from the last terman = l – (n-1)dan = 1/r(n-1)
Sum of first n termssn = n/2(2a + (n-1)d)sn = a(1 – rn)/(1 – r) if r < 1

sn = a(rn -1)/(r – 1) if r > 1

Sequence and Series Important Formulas

Difference Between Sequences and Series

Let us find out how a sequence can be differentiated with series.

SequencesSeries
Set of elements that follow a patternSum of elements of the sequence
Order of elements is importantOrder of elements is not so important
Finite sequence: 1,2,3,4,5Finite series: 1+2+3+4+5
Infinite sequence: 1,2,3,4,……Infinite Series: 1+2+3+4+……

Some Sequences and Series Solved Examples

Example 1: In a G.P, the second term is 12 and the sixth term is 192. Find the 11th term.

Solution:

Given second term, ar = 12 ..(1)

Sixth term, ar5 = 192 …(2)

Dividing (2) by (1) we get ar5 / ar = 192/12

⇒ r4 = 16

⇒ r = 2

Substitute r in (1), we get a×2 = 12

⇒ a = 12/2 = 6

11the term is given by ar10 = 6×210 = 6144

Hence the required answer is 6144.

Example 2: If Sn denotes the sum of first n terms of an A.P. and [S3n − Sn−1] / [S2n −S2n−1] = 31, then the value of n is __________.

Solution:

S3n = [3n / 2] [2a + (3n − 1) d]

Sn−1 = ([n − 1] / 2) * [2a + (n − 2) d]

⇒ S3n − Sn−1 = [1 / 2] [2a (3n − n + 1)] + [d / 2] [3n (3n − 1) − (n − 1) (n − 2)]

= [1 / 2] [2a (2n + 1) + d (8n− 2)]

= a (2n + 1) + d (4n2 − 1)

= (2n + 1) [a + (2n − 1) d]

S2n − S2n−1 = T2n = a +(2n − 1) d

⇒ S3n − Sn − S2n −S2n−1 = (2n + 1)

Given, ⇒ [S3n − Sn−1] / [S2n − S2n−1] = 31

⇒ n = 15

Example 3: If the sum of n terms of an A.P. is (c*n) (n – 1), where c ≠ 0 then the sum of the squares of this term is __________.

Solution:

If tr is the rth term of the A.P., then

tr = Sr − Sr−1 = cr (r − 1) − c (r − 1) (r − 2) = c (r − 1) (r − r + 2) = 2c (r − 1)

We have, t12 + t22+……..+tn= 4c2 (o2 + 12 + 22… + (n − 1)2)

= [4c2] * [(n − 1) n (2n − 1)] / [6]

= [2 / 3] c2 n (n − 1) (2n − 1)

Example 4: The largest term common to the sequences 1, 11, 21, 31,…. to 100 terms and 31, 36, 41, 46,………to 100 terms is ___________.

Solution:

100th term of 1, 11, 21, 31, …………. is 1+ (100 – 1) 10 = 991.

100th term of 31, 36, 41, 46, ………… is 31 + (100 – 1) 5 = 526.

Let the largest common term be 526.

Then, 526 = 31 + (n – 1) 10

Or n = 50.5

But n is an integer; hence n = 50.

Hence, the largest common terms are 31 + (n – 1) 10 = 521.

Example 5: If x, 2y, 3z are in A.P. where the distinct numbers x, y, z are in G.P., then the common ratio of the G.P. is ___________.

Solution:

x, y, and z are in G.P.

Hence, Y= xr, zy = xr, x = xr2

Also, x, 2y, and 3z are in A.P.

Hence, 4y = x + 3z ⇒4xr = x + 3xr2

⇒ 3r2 − 4r + 1 = 0

⇒ (3r − 1) (r − 1) = 0

⇒ r = 1/3

Sequences and Series Problems And Solutions

Question 1: Which number is wrong in the given series? 12439, 23549, 34659, 45769….

a) 34659

b) 23549

c) 45769

d) 12439

Answer: Option (A)

Explanation:

here we need to find the wrong number in 12439, 23549, 34659, 45769 as we will add the digits of every number

12439 : 1+2+3+4+9 = 19

23549 : 2+3+5+4+9 = 23

34659 : 3+4+6+5+9 = 27

45769 : 4+5+7+6+9 = 31
As we can see that in very case the sum of digits of number is a prime number except in case of 34659 and hence 34659 is the odd one out.

Question 2: Find the missing number: 2, 10, 20, 32, 46,?

a) 62

b) 65

c) 64

d) 60

Answer: Option (A)

Explanation:

let the missing number be y

10 – 2 = 8

20 – 10 = 10

32 – 20 = 12

46 – 32 = 14

as we can see that difference is getting increased by 2 and hence

y – 46 = 16

y = 62

Question 3: What will come in place of the question mark (?) in the series? 3, 8, 27, 112, (?), 3396.

a) 565

b) 452

c) 560

d) 678

Answer: Option (A)

Explanation:

Let the missing term be y

here the sequence is 3, 8, 27, 112, y, 3396

here the pattern is :

8 = 3 x 2 + 2

27 = 8 x 3 + 3

112 = 27 x 4 + 4

so from the highlighted pattern , we can say that

y = 112 x 5 + 5 = 565

 Question 4: If 1 + 4=9, 2 + 8=18, 3+6 = 15 then 7 + 8 = ?

a) 41

b) 23

c) 30

d) 32

Answer: Option (B)

Explanation:

Here we are given :

1 + 4=9

2 + 8=18

3+6 = 15

So if the number is , x + y then it is equal to [(x+y) + y]

here we need to calculate value of 7 + 8 = [(7 + 8) + 8] = 23

Question 5: Find the wrong number in the series from the given alternatives 17, 36, 53, 68, 83, 92.

a) 53

b) 68

c) 83

d) 92

Answer: Option (C)

Explanation:

The terms in the series are 17 + (17+19) + (17+19+17) + (17+19+17+15) + ( 17+19+15+15) + (17+19+15+13+11)

Each term adds the odd number which precedes the one added to the previous term, except for the fifth term.
The correct sequence would be 17,36,53,68,81,92.

Hence Option C is the correct answer.

Question 6: Find the wrong number in the series. 6, 9, 15, 22, 51, 99

a) 99

b) 51

c) 22

d) 15

Answer: Option (C)

Explanation:

9 -6 =3

15 -9 =6

22 -15 =7

51 -22 =29

99 -51 =48

So Ignoring the difference between the terms where one of the terms is 22 all other differences are multiple of 3

So 22 is the wrong term and the correct series would have been

9 -6 = 3 x 1

15 -9 = 3 x 2

27 -15 = 3 x 4

51 -27 = 3 x 8

99 -51 = 3 x 16

So 22 is the wrong term

 Question 7: Find out the set of numbers amongst the four sets of numbers given in the alternatives which is most like the set given in the question. Given Set : (8, 56, 72)…

a) (7, 56, 63)

b) (3, 15, 24)

c) (6. 42, 54)

d) (5, 30, 35)

Answer: Option (C)

Explanation:

(8, 56, 72)

follows the pattern

(8 , 8 x 7 , 8 x 9 )

So (6. 42, 54) is similar to (8, 56, 72)

as it follows the pattern

(6 , 6 x 7 , 6 x 9 )

Question 8: Which of the given letter groups follows the same principle on which the numbers are arranged in the given number group ? (5, 8, 13, 20, 29)

a) ACGKR

b) ADIMS

c) ADIPY

d) ADGRO

Answer: Option (C)

Explanation:

The Series follows the pattern

5

5+3 =8

8+5 =13

13 +7 =20

20+9 = 29

So the option representing the correct letter group must follow the same pattern

So the letter series is

A

A+3 = D

D +5 =I

I +7 = P

P + 9 =Y

So the letter series is ADIPY

 Question 9: Find out the wrong number in the sequence 102, 101, 98, 93, 86, 74, 66, 53

a) 101

b) 66

c) 74

d) 93

Answer: Option (C)

Explanation:

The Series follows pattern

102

102 -1 =101

101 -3 =98

98 -5 =93

93 -7 =86

Since 1,3,5 and 7 are subtracted in order to get the next term

So 7 +2 =9

should be subtracted in order to get next term.

So next term should be

86 -9 =77

77 -11 =66

66-13 =53

So option C is correct

Question 10: Based on the given data, estimate the number of ‘Television buyers’ for the year 1990.

1982 1984 1986 1988 1990

447 458 489 540 ?

a) 611

b) 591

c) 571

d) 601

Answer: Option (A)

Explanation:

Number of television Buyers in 1982 = 447

Number of Television buyers in 1984 = 458 (447+11)

Number of Television buyers in 1986 = 489 (458+ 11 +20)

Number of Television buyers in 1988 = 540 ((489 + 11 +20 +20 )

Number of Television buyers in 1990 = 611 (540 + 11 +20 +20 + 20 )

Download Sequences And Series Question with Solution PDF

Note:

Students must practice out Sequence and Series questions which will help you to clear your basic concepts and it will help you in improving your performance. We hope that the Sequence and Series Tricks given here on this page of recruitmentresult.com will be useful for the students going to appear in the upcoming examinations.

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