In this section, we are going to explore **Sequence and Series** and this is the easiest and important concept.

What is the sequence?

A sequence is a set of numbers in which numbers occur in a definite order or governed by a rule.

For a simple understanding when we are in an unknown colony to find a specific house number. We simply notice that house numbers are in order e.g. (61,62,63…), Yes it is a sequence of house numbers.

**PROGRESSIONS**:

Sequences of special types are called progressions where the terms of the sequence follow a particular pattern.

**TYPES OF PROGRESSION**

- Arithmetic Progression
- Geometric Progression
- Harmonic Progression

In this unit, we will discuss AP and HP.

**ARITHMETIC PROGRESSION**:

A sequence a_{1}, a_{2}, a_{3} … a_{n} where a_{2} – a_{1} = a_{3} – a_{2}

= a_{n} – a_{n-1}, Where quantities increased or decreased continuously with a common quantity.

E.g. 1,3,5,7,9,…

Common difference = 2

First term = 1 and General term (nth term) of A. P.

There is a rule binding all the terms of progression so we could tell more about the terms by following that rule.

By going though previous example we saw

⇒First term = 1

⇒Second term = 3 = 1 + 2 (one times 2)

⇒Third term = 5 = 1 + 2 + 2 (two times 2)

⇒Fourth term = 7 = 1 + 2 + 2 + 2 (three times 2)

Simmilarly

⇒nth term will be = 1 + 2 + 2 … (n – 1 times 2)

So nth term would be = first term + (n – 1) × difference

Hence

T_{n} = a + (n – 1) × d

Where

T_{n} = nth term of A.P.

a = First term of A.P.

n = Number of terms in a progression

d = Difference between terms

E.g. Find the 17th term of -7, -2, 3, 8, 13, …

Here

n = 17

a = -7

d = -2 – (-7) = 5, 3 – (-2) = 5

So according to formula

⇒ T17 = a + (n – 1) × d

⇒T17 = -7 + (17 – 1 ) × 5

⇒T17 = -7 + 16 × 5

⇒T17 = -7 + 80

⇒T17 = 71(Ans.)

- Sum of n terms of A.P.

⇒ Sn = n / 2 × [2a + (n – 1) × d]

We can write it as

⇒ Sn = n / 2 × [a + an]

E.g. Find the sum of 11 terms of -7, -2, 3, 8 …

We noticed that the terms are A.P.

So by the sum would be

Sn = n / 2 × [2a + (n – 1) × d]

Where

n = 11

a = -7

d = 5

By substituting the values

S_{n} = 11 / 2 × [2 × (-7) + (11 – 1) × 5]

S_{n} = 11 / 2 × [-14 + 10 × 5]

S_{n} = 11 / 2 × [-14 + 50]

S_{n} = 11 / 2 × [36]

S_{n} = 11 × 18

S_{n} = 198 (Ans.)

Special Point: When three quantities are in A.P. then the middle one is called the arithmetic mean of the other two terms.

A.M. = (a + b) / 2

E.g. If the sum of the first five terms of an A.P. is 75, then find the third term of the A.P.

Let the terms of A.P. be

a – 2d, a – d, a, a + d, a + 2d respectively.

According to question sum would be

⇒ a – 2d + a – d + a + a + d + a + 2d

⇒ 5a = 75 (Given)

⇒ a = 15 (Ans.)

As we know the third term is ‘a’ so our answer would be 15.

TIP- If we have to let terms in A.P. then we should assume the middle term ‘a’ and the term before them as ‘a – d’, ‘a – 2d’… and terms after them should be ‘a + d, ‘a + 2d’ … so we could cancel out terms easily.

**Hamonic Progression**:

A sequence of quantities a_{1}, a_{2}, a_{3}… an is said to be H.P. when their reciprocals are in A.P.

⇒ 1 / a_{2} – 1 / a_{1} = 1 / a_{3} – 1 / a_{2} = 1 / a_{4} – 1 / a_{3}

For example a sequence of terms like

1, 1 / 2, 1 / 3, 1 / 4 …

There is no general formula for H.P. but we can solve these questions by inverting the terms and using properties of A.P.

Harmonic Mean: When three quantities are in H.P. the middle one is called the Harmonic Mean of the other two.

E.g. Find 10th term of H.P. 1, 1 / 3, 1 / 5, 1 / 7, 1 / 9,…

So by inverting the progression we saw an A.P. 1, 3, 5, 7, 9…

It’s 10th term would be

⇒ T10 = a + (n – 1) × d

⇒ T10= 1 + (10-1) × 2

⇒ T10 = 1 + 9 × 2

⇒ T10 = 1 + 18 = 19

By inverting 19 we got 1 / 19 so our answer is 1 / 19.

In the above section we have learned about A.P. and H.P. in sequences, in this unit we will get an idea about G.P. (Geometric Progression) and explore about different types of series.

So let’s talk about G.P. first, what is G.P., which type of questions to be asked in G.P.?

__Geometric Progression__**:**

A sequence a_{1}, a_{2}, a_{3 }… a_{n} is said to be in G.P. when

a_{2} / a_{1 }= a_{3} / a_{2} = a_{4} / a_{3} … a_{n} / a_{n-1} = r

Where a_{1}, a_{2}, a_{3 }… a_{n} are non-zero numbers and ‘r’ is said to be common ratio of the G.P.

**E.g. **2, 4, 8, 16, 32…

We saw in this example that as the first term is 2 then the next term would be 2 × of that

Similarly, the third term is twice the second term so it is a G.P.

**General Term (nth term) of G.P.**

The general term of a geometric progression is

T_{n} = a.r^{n-1}

Where

T_{n} = nth term

a = First term

r = Common ratio

n = Number of terms

- If the common ratio (r) of G.P. is greater than 1 than the terms strictly increase. So, we can see

a_{n} > a_{n-1 }> …a_{4} > a_{3} > a_{2} >a_{1}.

- Similarly, when common ratio (r) of G.P. is less than 1 than the terms strictly decreases. So, we can see

a_{n} < a_{n-1 }< … a_{4} < a_{3} < a_{2} <a_{1}.

- When common ratio (r) = 1, then all the terms would be equal

an = an-1 = … a4 = a3 = a2 = a1.

**Sum of terms of G.P.**

- When the common ratio > 1

S_{n} = [ a(r^{n} – 1) ] / (r – 1)

- When the common ratio < 1

S_{n} = [ a(1 – r^{n}) ] / (1 – r)

- When common ratio = 1

S_{n} = n × a

Now w will try to solve some questions related to it.

**E.g.** Find 8^{th}, 11^{th} term of G.P. and find the sum of the first 11 terms.

G.P. ⇒ 3, 6, 12, 24, 48,…

In this sequence we saw that the difference of terms is not equal : 6 – 3 ≠ 12 – 6

But we can saw another type of rule here as

6 / 3 = 12 / 6 = 24 /12 = 48 / 24 = 2

Here the ratio of terms is equal it is a G.P.

From the nth term formula

⇒ T_{8} = a.r^{8-1}

⇒ T_{8} = 3 × 2^{7 }= 3 × 128

⇒ T_{8} = 384 (**Ans**.)

Similarly

⇒ T_{11} = a.r^{11-1}

⇒ T_{11} = 3 × 2^{10 }= 3 × 1024

⇒ T_{11} = 3072 (**Ans**.)

From the formula of the sum of ‘n’ terms

Sum of first 11 terms would be = [ a(r^{n} – 1) ] / (r – 1) (r > 1)

S_{11} = [3 × (2^{11} – 1)] / (2-1)

S_{11} = [3 × (2048 – 1)] / 1

S_{11} = [3 × 2047]

S_{11} = 6141 (**Ans**.)

**E.g.** 256, 192, 144, (?)

Here the difference is not equal so let’s check the common ratio

192 / 256 = 144 / 192 = 3 / 4

OHH so it’s a decreasing G.P. with common ratio 3 / 4

So fourth term would be T_{4} = a.r^{4-1}

⇒ T_{4} = 256 × (3 / 4)^{3}

⇒ T_{4 }= 256 × 27 / 64

⇒ T_{4} = 108 (**Ans**.)

**Special point:** If the three quantities are in G.P., the middle one is called the geometric mean (G.M.) of the other two quantities.

If a, m, b are the terms of G.P.

m = √ab OR m^{2} = a × b

__SERIES__

A number series is a sequence of numbers written from left to right in a certain pattern. To solve the questions on series, we have to detect/find the pattern that is followed in the series between the consecutive terms, so that the wrong/missing term can find out.

**Infinite Series**

Firstly we will learn about an infinite series, where terms have a common ratio which would be less than 1.

**E.g. **1 + 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 …

Here the common ratio is 1 / 2 ÷ 1 = 1 / 2

The formula for sum of infinite terms is = a / (1 – r)

So in the above question

⇒ S = a / (1 – r)

⇒ S = 1 / (1 – 1 / 2)

⇒ S = 1 / (1 / 2)

⇒ S = 2 (**Ans**.)

**Prime Number Series:**

The number which is divisible by 1 and itself, is called a **prime number.**

First 5 prime numbers are 2, 3, 5, 7, 11

**E.g. **61, 67, 71, (?), 79

There is neither constant difference between terms nor they are in fix ratio so what would we do.

We saw all terms are Prime numbers here, what would be next prime number?

It is 73 (**Ans**.)

In the quant section we learned many concepts of **in the above section**, In this secction, we will be going to study different types of series.

So stop wasting time let’s gear up and start with first one;

1. **n ^{2} Series**

When a number is multiplied with itself, then it is called a square of a number, and the series formed by the square of numbers is called the *n ^{2} *series. Here some square is added to the previous term OR some square is subtracted from the previous term.

**E.g.** Find the missing item in the following series.

20, 29, 54, 103, 184, (?)

**APPROACH: **What should be our first response after viewing this series? Right now we are completely clueless. The first thing we noticed is it is increasing series so some term is added to the previous term okay.

Let’s see the difference between the terms.

29 – 20 = 9 = 3^{2}

54 – 29 = 25 = 5^{2}

103 – 54 = 49 = 7^{2}

184 – 103 = 81 = 9^{2}

It is a series where the square of odd numbers beginning from 3 is added to previous terms, after 9^{2} we should square the next odd number 11 that is 121

⇒ 184 + 121 = 305 (**Ans**.)

2. **( n ^{2} ± 1) Series**

If in a series each term is an addition/subtraction of a square term and 1, then this series is called (n^{2} ± 1) series.

⇒ 1^{2} ± 1, 2^{2} ± 1, 3^{2 }± 1

**E.g.** Find the missing item.

1254, 1153, 1088, (?), 1034

**APPROACH: **Our mind should be trained like that if we didn’t saw a direct series then instantly we should find the difference of terms.

1254 – 1153 = 101

1153 – 1088 = 65

We can see that 101 = 10^{2} + 1 & 65 = 8^{2} + 1 next term would be 6^{2} + 1 i.e. 37.

⇒ 1088 – 37 = 1051 (**Ans**.)

We can even cross-check our answer by checking the difference between 1051, 1034

It should be 4^{2} + 1 = 17 = 1051 – 1034

Yes, we got the right answer.

See cracking competition exams is a skill you have to practice the skill to make yourself perfect. Let’s take the next type of series.

3. **(n ^{2} ± n) series**

Those series in which each term is an addition or subtraction of a number with the square of that number is called as *(n ^{2} *±

**E.g**. 420, 930, 1640, (?), 3660

Here we see terms are increasing with pace so we can assume there might be any series of type n^{2}, n^{3,} or higher powers.

Here by calculating difference we can’t find any specific sequence but if we look

⇒ 420 = 400 + 20 = 20^{2} + 20

⇒ 930 = 900 + 30 = 30^{2} + 30

⇒ 1640 = 1600 + 400 = 40^{2} + 40

Similarly next term would be 50^{2} + 50 =2500 + 50 =2550 (**Ans**.)

**E.g. **210, 240, 272, 306,342, …

Here noticing terms clearly, we can see

210 = 15^{2} – 15

240 = 16^{2} – 16

272 = 17^{2} – 17

306 = 18^{2} – 18

So we saw there is a a sequence of ‘n^{2} + n’

342 = 19^{2} – 19

next term would be = 20^{2} – 20 = 380(**Ans**.)

OR

The difference between terms is 30, 32, 34, 36 so we saw it is forming an A.P. next term would be 38

⇒ 342 + 38 = 380 (**Ans**.)

What we saw is that to reach a solution we can find different ways and still we can reach the right answer if the series is followed by all terms.

4. **n ^{3} Series**

A number when multiplied with itself twice, then the resulting number is called the cube of a number and series which consist of cube of different number following a specified sequence, is called as n^{3} series.

**E.g. **8850, 5475, 3278, 1947, 1218, (?).

8850 – 5475 = 3375 = 15^{3}

5475 – 3278 = 2197 = 13^{3}

3278 – 1947 = 1331 = 11^{3}

1947 – 1218 = 729 = 9^{3}

Next difference between terms will be 7^{3} = 343

1218 – (?) = 343

(?) = 1218 – 343 = 875 (**Ans**.)

5. **(n ^{3} ± 1) Series**

Those series in which each term is a addition or subtraction of a cube of a number and 1, are known as *(n ^{3} *+1) series.

**E.g.** 124, 213, 338, (?), 720, 989

124 = 125 – 1 = 5^{3} – 1

213 = 216 – 3 = 6^{3} – 3

338 = 343 – 5 = 7^{3} – 5

(?) = 8^{3} – 7 = 512 – 7 = 505 (**Ans**.)

**6. Alternating Series**

In alternating series, successive terms increase and decrease alternately. The possibilities of alternating series are

♦ If It is a combination of two different series.

♦ Two different operations are performed on successive terms alternately.

**E.g.** Find the next term in the series 15,14,19,11, 23, 8, …

So simply 23 + 4 =27 (**Ans**.)