**Number Sequence:**Number of sequence is an ordered list of numbers.

**For example:**

1,3,5,7,9,11

**.......**

2,4,6,8,10,12......

2,4,8,16,32,63........

**Arithmetic progression:**

When a constant of number is added/ subtracted from a number to form the next number in the sequence, it is called an arithmetic progression. eg. 2,4,6,8.... is an arithmetic progression where we can add 2 to a number and get the next number in the sequence.

**n-th term:**

Let a be the 1st term and d be the 'common difference'

**.**So, the 2nd term is (a+b)

**,**3rd term is (a + 2d) ,.....and so no. So in a similar fashion the n-th term will be [a+n(n-1)d] .

**Summation of the first n terms of an arithmetic progression:**

Often you will be asked to find out the summation of the first n terms of an arithmetic progression. The formula for this is

**1/2[n+(n-1)d]**

**where a = 1st term; d= common difference.**

a very important special case is the series

1+2+3+4+5+...................+n

Here a= 1, d=1

:. Summation of n terms = 1/2[n{2.1+(n-1).1}]

= 1/2{n(2+n-1)}

= 1/2{n(n+1)}

This is a very useful formula. So, please memorize it.

Often in tests you will find the term 'consecutive numbers. Consecutive numbers are only a special case of arithmetic progression. For example, the consecutive integer of 5 is 6.

The following is a sequence of consecutive integers

5,6,7,8,9,10 ........

Again, the following sequence is a sequence of consecutive even numbers.

2,4,6,8,10.......

Similarly, 7,9,11,13,15....... is sequence of consecutive odd numbers. In the solved example's section & in the exercises, you will find many problems related to consecutive numbers.

**Geometric progression:**

In a sequence of numbers if each term may be multiplied / divided by a constant number (the 'common ratio') to give the next number in the sequence, it is called a geometric progression. For example, 2,6,18,54,162....... is a geometric progression where each term is multiplied by 3 to produce the next term in the sequence.

**n-th term of the series:**

The first term is usually denoted by a and the common ratio by r. So, the n-th term is given by

*ar*

^{ n-1}**.**

**Summation of the first n terms of a geometric progression:**

The sum of the first n terms of a geometric progression is

**{a(1-**

*r*)}/(1-r)

^{ n}Now, let us go through a few examples.

**1. The seventh term of an arithmetic progression is 56 and the**

*11*

^{ th}**term is 88. Calculate the 20**

^{ th}**term & the sum of the first 1000 terms.**

Solved: Let, a be the 1

^{th}**term & d be common difference. Hence,**

7

*term= a + 6d = 56..........................(i)*

^{ th}11

*term = a+ 10d = 88 ....................(ii)*

^{ th}(ii) - (i) given 4d = 32, d= 8

:. a+6x8 = 56=a =56-48=8

:. 20

*terms = a + 19d = 8+19x8*

^{ th}= 160

Sum of the 1

*100 terms = 1/2[n{2n+(n-1)d}]*

^{ th}= 1/2[100{2.8+99.8}]

= 40400

**2. An investor starts with $ 500 in an investment account and each month it earns a constant**$

**32 interest. After how many years will the sum in the account exceed $ 700?**

Solved: In this problem we will use the n-th term formula. Because every month $32

**is added with teh previous month's balance to produce the next month's balance.**

:. a = 500, d=32

So, accouding to the problem.

a+(n-1)d= 700

or 500 + (n-1).32= 700

or (n-1).32 =700

or, 32n = 200 + 32= 232

:. n= 232/32 = 7.25 months

So, after 8 months, the balance will exceed 700.

Usually , MBA questions & banks you are asked to find the next number in a sequence. For example, you may be given the following: 2,4,6,8 and asked to find out the next number.

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