# Reasoning - Sequence and Series

## Number Sequence and Series

**Sequence: **A sequence is an ordered list of numbers** or a** sequence is a set of numbers arranged in a particular order.

**Series: **A series is composed of a sequence of terms that are added up.

Let us start with few simple examples of number sequences.

- Example 1: 1,2,3,4,5,6,7,8,....
- Example 2: 7,10,13,16,19,22,.....
- Example 3: 12,17,22,27,32,37,.....

- Example 4: 1,2,4,8,16,32,64,128,.....
- Example 5: 3,9,27,81,243,749,2247,.....
- Example 6: 1/2,1/4,1/8,1/16,1/321,1/64,....

- Example 7: 1,1/2,1/3,1/4,1/5,1/6,1/7,1/8,....
- Example 8: 1/7,1/10,1/13,1/16,1/19,...
- Example 9: 1/12,1/17,1/22,1/27,....

Above mentioned examples are mainly in three mathematical serquences. They are AP, GP, and HP. AP stands for arithmetic progression, GP stands for geometric progression, and HP stands for harmonic progression. First three (1,2,3) examples are in AP, second three examples (4,5,6) are in GP, and the last three examples (7,8,9) are in HP. Let us try to understand the three sequences one by one.

## Arithmetic Progression (AP):

Arithmetic Progression is also called arithmetic sequence. In a sequence of AP, the difference between any two consecutive numbers or terms is constant and this constant, we are denoting as d, is called common difference. If first or last term of the series is known and we know the common difference, we can get the next term or any term of the series. Even, we can calculate the sum of, say n number of terms. Many questions related to AP are asked in examination. Let us explain in detail with an example.

An arithmetic progression is given by the formula:

a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d),+.......[a+(n-1)d], ......

where a = first term, d = common difference

**Examples**

4, 6, 8,10,12,14,..... is an arithmetic progression with a = 4 and d = 2

13,18, 23, 28, 33,.... is an arithmetic progression with a = 13 and d = 5

**Note: **If the three terms a, b, c are in AP then,

b = (a+c)/2

**or**

2b = a + c

**n ^{th} term of an arithmetic progression:**

a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d),+.......+ [a+(n-1)d], .....

is given by the formula:

.................................

t_{n} = a + (n – 1)d = l

.................................

where t_{n} = n^{th} term, a = the first term , d = common difference, l = the last term

**Sum of n terms in an arithmetic series:**

**.................................................................**

S_{n} = n/2 [ 2a + (n-1)d ] = S_{n = }n/2( a + l )

**.................................................................**

where, a = first term of the series

n = number of terms in a series

d = common difference

t_{n =} n^{th }term of the series

S_{n} = Sum of n terms of the series

l = last term of the series

**To find the sum of an AP series when the first term (a),the common difference (d), and the number of terms (n) are known.**

**Example1:**

Find sum of the first 100 terms of the series?

1 + 2 + 3 + 4 + 5 + 6+............................100 + 101 +.....

**solution:** Here, a = 1, d = 1, n = 100, so

S_{100} = 100/2[ 2×1 + (100 - 1)×1 ]

S_{100 }= 50[ 2 + 99 ]

S_{100 }= 50[101]

S_{100} = 5050

**To find the sum of an AP series when the first term (a), the common difference (d), and the last term (n) are known.**

**Example 2:**

Q. Find the sum of the series:

5 + 10 +15 +20 +25 +............+60

**Solution**: Here, a = 5, l = 50, d =10 - 5 = 20 -15 = 5 (take any two consecutive terms, you will get the same difference)

n = ?, S_{n = ?}

To fins the sum first, we will have to find n (number of terms)

Using formula,

l = a + (n - 1)d

60 = 5 + (n - 1)5

55 = 5n - 5

5n = 55+5

5n = 60

n = 12 (dividing both the sides by 12)

Now, using sum formula

Sn = n/2[a + l]

Sn = 12/2[5 + 60]

Sn = 6[65]

Sn = 390

**Arithmetic Mean**

If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c.

In this case,

b = (a+c)/2

2b = a+c

**Note**: If a, a_{1}, a_{2} ... a_{n}, b are in AP we can say that a_{1}, a_{2}, a_{3}, .... a_{n} are the n Arithmetic Means between a and b.

## Geometric Progression:

Geometric Progression (GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant. In other words, it is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (r).

A geometric progression (GP) is given by the formula:

a, ar, ar^{2}, ar^{3}, ...ar^{n−1},.....

where a = the first term, r = the common ratio, n = number of terms.

**Examples**

- 3, 6, 12, 24, 48, ... is a geometric progression (GP) with a = 3 and r = 2
- 1, 2, 4, 8, 16, ... is a geometric progression (GP) with a = 1 and r = 2

**Geometric Mean:**

** **If a, b, c are in GP then,

b^{2} = ac

or

.............

b = √ac

.............

where b is the geometric mean between a and c

**n ^{th} term of a geometric progression (GP):**

**a, ar, ar ^{2}, ar^{3}, ...ar^{n−1},.....**

**...............**

t_{n }= ar^{n−1}

**...............**

where t_{n} = n^{th} term, a = the first term , r = common ratio, n = number of terms

**Sum of n terms of a geometric series:**

Let a geometric series be represented by the formula:

a + ar + ar^{2 }+ ar^{3} + ar^{4 }+ .........ar^{n-1}......, where a = first term and r = common ratio, then

**Example 1**

Find the 10^{th} term in the series 2, 4, 8, 16, 32,....

a = 2, r = 4/2 = 2, n = 10

10^{th} term,

t_{10 }= ar^{n−1}=2×2^{10−1}= 2×2^{9 }= 2×512 =1024 →Ans_{.}

**Example 2**

Find 6^{th} term in the series 5, 15, 45, 60, ...

a = 5, r = 15/5 = 3, n = 6

6^{th} term,

t_{6 }= ar^{n−1 }= 5×3^{6}^{−1 }= 5×3^{5}= 5×243 = 1215 →Ans_{.}

### Sum of Infinite Geometric Series

A finite sum can be obtained from GP with infinite terms if and only if -1.0 ≤ r ≤ 1.0 and r ≠ 0.

## Harmonic Progression

The reciprocals of the corresponding terms of an AP, containing non-zero terms, constitute an HP i.e. if 1, 2, 3, 4, 5, ....... are in AP, then 1, 1/2, 1/3, 1/4, 1/5,..... are in HP. But if the sequence is 0, 1, 2, 3 ,4, 5 ..... , the reciprocals of its corresponding terms will not constitute an HP. Reciprocal of zero is not defined.

**n**^{th }** term of an H.P.**

If a, a + d, a + 2d, a + 3d,............ a + (n-1)d are in AP,

where n^{th} term of AP = a + (n - 1)d,

then 1/a, 1/(a + 2d),1/(a + 3d),........1/[a + (n-1)d] are in HP.

where,

**.................................................**

n^{th} term of HP = 1/[a + (n-1)d]

**.................................................**

** Examples:**

**Q.1 **Find the 16^{th } term of the given HP.

1, 1/3, 1/5, 1/7,.......

**Solution:** Using formula,

t_{n} = 1/[a + (n - 1)d]

t_{16} = 1/[1 + (16 - 1)2]

t_{16 = }1/[1 + 30]

t_{16 = }1/31 →Ans_{.}

**Harmonic Mean**: If a, b, c are in HP and a ≠ 0, b ≠ 0, c ≠ 0, then

1/a, 1/b, 1/c are in AP

→ 1/b = (1/a + 1/c)/2

**...............................................**

→ b = 2ac/(a + c)

**...............................................**

where b is the harmonic mean between a and b

## Try, what you have learned.

Just go through this quiz and judge yourself.

## Number Series test 1

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