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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 n )}/(1-r)
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 1th term & d be common difference. Hence,
7 th term= a + 6d = 56..........................(i)
11 th term = a+ 10d = 88 ....................(ii)
(ii) - (i) given 4d = 32, d= 8
:. a+6x8 = 56=a =56-48=8
:. 20 th terms = a + 19d = 8+19x8
= 160
Sum of the 1 th 100 terms = 1/2[n{2n+(n-1)d}]
= 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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