*Friday, September 22nd, 2017*

Sequence and Series topic of Quantitative Aptitude is one the most engaging and intriguing concept in CAT. Since childhood, we love solving puzzles based on sequence and series. Sequence and series are closely related concepts and possess immense importance. We all have heard about the famous Fibonacci Sequence, also known as Nature’s code. This sequence can be used to model or describe various phenomena in the field of biology, mathematics, nature, art etc. The sequence is **0, 1, 1, 2, 3, 5, 8, 13… so on. **As we can see, the next no. in the sequence is the addition of previous two numbers in the sequence. This is just one very significant example of sequences. In an equivalent way, there can be thousands of sequence and series that can be formed using a just various combination of these numbers beginning from 0 to 9. Isn’t that amusing!

Today, in this blog, we will learn about one specific branch of sequence and series i.e. Progression and its related concepts. Let’s begin first with formally defining sequence, series, and their difference.

A sequence is the list of numbers written in some specific order. For example, etc.

And, series is what we obtain by adding up the terms of the sequence. For example, etc.

Now let’s finally come to Progression, they are set of numbers which are arranged according to some definite rule. The major difference between sequence and progression is that progression has a specific formula to calculate its nth term whereas sequence can be based on any logical rule. So, every progression is a sequence but not every sequence is a progression.

Progression is of three types:

- Arithmetic Progression.
- Geometric Progression.
- Harmonic Progression.

It is a sequence of numbers in which each subsequent term of it is obtained by adding up a constant term to it. That is, the difference between two consecutive terms remains same throughout. This constant term is known as **Common Difference **and is denoted by ‘**d**’ and the first term of the sequence is denoted by ‘**a**’. This progression is of form **a, a+d, a+2d, a+3d… **For instance, consider the sequence **1, 4, 7, 10, 13, 16. **You can notice that in this example the difference between any two terms of the sequence is 3 (4-1=3, 7-4=3, 10-7=3 and so on.) Therefore, in this AP the common difference ‘d’ is 3 and first term ‘a’ is 1.

Suppose we have another A.P. with the first term as 1 and common difference as 5 and we need to calculate the 7^{th} term of this sequence. How to find this out?

Now the sequence can be formed as 1, 6, 11, 16… and we are interested to get the 7^{th} term of this sequence. We can write it in the form a + 6d and thus get 31. Generalizing it, we get a + (n-1)d. Hence, the formula we get for nth term of A.P. denoted by **T _{n }= a+ (n-1)d**.

What if you have to calculate the sum to n terms of the above sequence then you can use the formula given as **S _{n }= n/2 {2a + (n-1) d} or n/2 {a + l}, **where l is the last term of A.P. Thus, using this formula we get S

Let’s now consider another comprehensive example that will highlight another important concept of A.P.

The sum of first three no. of A.P. is 27 and the sum of their squares is 293. Then find the numbers.

To devise the solution to above problem we cannot directly take the three terms as a, a+d, a+2d and then with the help of the T_{n }and S_{n} formula we calculate the numbers. This approach will take you nowhere. Instead, you can solve it using the following way.

Assume the three terms to be a-d, a, a+d. Since, the sum of first three terms is 27. So,

a-d + a+ a + d = 27

Cancelling out d we get,

3a = 27

a = 9

Also, (a-d)² + a² + (a+d)² = 293

243 + 2d² = 293

2d² = 50

d² = 25

d = ±5

When d= -5, the no.’s is 14, 9, 4

When d= 5, the no’s are 4, 9, 14.

Thus, by using this approach we can quickly and easily determine the three numbers. Likewise, if terms are more than 3 then follow up the table given below.

No. of terms |
Terms |
Common Difference |

4 |
a-3d, a-d, a+d, a+3d | 2d |

5 |
a-2d, a-d, a, a+d, a+2d | d |

6 |
a-5d, a-3d, a-d, a+d, a+3d, a+5d | 2d |

**Relationship between Arithmetic Mean and A.P.**

**Let A be the arithmetic mean of a and b. Then, a, A, b are in A.P. Therefore A – a = b – A => 2A = a + b = (a + b)/ 2**

This is a kind of sequence where each term after the first is obtained by multiplying the previous no. by a fixed constant number, known as the common ratio. It is denoted by ‘**r**’. It is of form a, ar, ar², ar³… The n^{th} term of the G.P. is T_{n} = ar^{(n-1)}_{. }For example, 2, 4, 8, 16, 32…so on is a G.P. with common ratio = a_{k+1}/a_{k} = 2.

To calculate the sum to n terms of G.P. of the above example use this formula.

S_{n} = a(1-r^{n}/ 1-r) = a/1-r -ar^{n}/1-r

And when n →∞ then r^{n} →0

Thus, sum of an infinite G.P. is given by

S = a/1-r

Where r is common ratio and a is the first term of G.P.

Suppose, 1/2^{3}, 1/2^{6}, 1/2^{9}…is an infinite G.P. Then its sum will be

S = 1/2^{3}/ (1 – 1/8) = 1/7

As given in the above example of A.P. in case of selection of terms similar kind of questions can also be asked for G.P. where you have been provided the sum of first 3 or 4 terms of G.P. and you need to find out the numbers. You can refer the table given below for the same.

No. of terms |
Selection of terms |
Common difference |

3 |
a/r, a, ar | r |

4 |
a/r^{3}, a/r, ar, ar^{3} |
r^{2} |

5 |
a/r^{2}, a/r, a, ar, ar^{2} |
r |

6 |
a/r^{5}, a/r^{3}, ar, ar^{3}, ar^{5} |
r^{2} |

Before moving forth let’s do another example of G.P. such as

The sum of first two terms of G.P. is 2 and the sum of first four terms of G.P is 20. Find the G.P.

Firstly, note that this question is not based on a selection of terms and here we won’t take terms of G.P. to be a/r^{3}, a/r, ar, ar^{3}.

In this question, we will proceed simply by taking terms to be a, ar, ar^{2}, ar^{3}.

Since, a + ar = 2 => a (1 + r) = 2

Also, a+ ar +ar^{2}+ ar^{3} = 20 => a (1 + r) (1 + r^{2}) = 20

⇒2(1 + r^{2}) = 20

⇒(1 + r^{2}) = 10

⇒r^{2} = 9

⇒r = ±3

⇒a = -1, when r = -3

⇒a = ½, when r = 3.

**Relationship between Geometric Mean and G.P.:**

Let G be the geometric mean between a and b. Then a, G, b are in G.P. And, G = √ab.

Harmonic Progression

It is a progression that is formed by taking reciprocal of arithmetic mean. It is of the form 1/a, 1/(a+d), 1/(a+2d) … For example, 1, 3, 5, 7, 9 forms an A.P. with common difference = 2 then H.P. will be 1, 1/3, 1/5, 1/7, 1/9.

Consider an example, If the sum of first 2 terms of H.P. is 17/20, the sum of next two terms of H.P. is 5/4, the sum of following two terms is -7/10. Find the sum of following two terms again.

In this question, we are given

1/a + 1/(a+d) = 17/20

1/(a+2d) + 1/(a+3d) = 5/4

1/(a+4d) + 1/(a+5d) = -7/10

Solving all 3 equations simultaneously we get, a = 10, d = -3.

Therefore, 1/(a+6d) +1/(a+7d) = -19/88.

**Relationship between Harmonic Mean and H.P.**

Let H be the harmonic mean between two non-zero numbers. Then a, H, b are in H.P. where H = 2ab/(a+b). So, 1/a, 1/H, 1/b are in A.P.

Now there could be hundreds of different questions based on the above progression. The questions that will come in exam will not be very simple and direct question based on the concept. It may be based on the combination of A.P and G.P, etc. and can also demand the knowledge of other concepts of math’s such as trigonometry, logarithmic, geometry, etc. You can be clueless how to proceed in the exam if you do not thoroughly prepare this concept. Therefore, you need to practice many lot of questions to develop and learn the application of progressions. As “** Practice creates confidence and confidence empowers you**”.

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