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Sequence and Series

A set of numbers arranged in a definite order according to some rules is called a sequence or progression. It means that all consecutive terms must be related by some common rule or property.

 

Example: 1, 2, 3, 4, 5… is a sequence of consecutive natural numbers.

 

Similarly, 2, 4, 6, 8, 10…. is a sequence of consecutive even numbers.

 

The expression of the sum of a sequence is called a series.

 

Example: 1 + 2 + 3 + ... is a series of natural numbers.

Arithmetic Progression

A sequence is called an arithmetic progression (A.P.) if the difference between any term and its preceeding term is a constant. This constant is called common difference of the A.P.

 

The common difference is found by subtracting any term of the sequence from the term next to it.

 

Example: 3, 5, 7, 9 ... is in arithmetic progression.

 

Here, 5 – 3 = 7 – 5 = 9 – 7 = 2 is the common difference.

 

Let ‘a’ be the first term of an A.P. and ‘d’ be the common difference, then the A.P. can be written as:

 

aa + da + 2da + 3d… and so on.

 

Now, we can see that the coefficient of ‘d’ is 1 in the 2nd term, 2 in the 3rd term, 3 in the 4th term. It is one less than the number of the term. So, for the nth term, coefficient of ‘d’ will be (n – 1).

 

Hence, we can write the nth term of an A.P. as:

 

Tn = a + (n – 1)d

 

Example
Form an A.P. where the 1st term is 5 and the common difference is 7.
Solution
Using the 1st term ‘a’ and the common difference ‘d’, the A.P. can be formed in the following way,
aa + da + 2da + 3d, … a + (n – 1)d
Here, a = 5 and d = 7
Substituting the values, the required A.P. will be
5, 5 + 7, 5 + 2(7), 5 + 3(7), … 5 + (n – 1)7 = 5, 12, 19, 26, … 5 + (n – 1)7

 

Example
Find the 10th term of the progression 3, 11, 19, 27…
Solution
From the given A.P., we can see that, a = 3, d = 8 and n = 10.
The nth term of an arithmetic progression is given by,
Tn = a + (n – 1) d
Substituting the values of the terms in the above relation, we get
T10 = 3 + (10 – 1)8 ⇒ T10 = 75

 

Example
Find the 6th term of the A.P. 5, 10, 15, 20.
Solution
Given n = 6, a = 5 and d = 5.
 
 
 

 

Example
If 6, 4m and 5m are three consecutive terms of an A.P., find ‘m’.
Solution
Given 6, 4m and 5m are three terms in A.P.
 

 

Example
Find the common difference, if T3 = 11 and T5 = 21.
Solution
Given T3 = 11 and T5 = 21.
 

Sum of Terms in an Arithmetic Progression

The sum of ‘n’ terms in an arithmetic progression is given by,

 

 

If we substitute the value of Tn, which is ‘a + (n – 1)d ’, in the above equation, we get

 

 

Where, a = 1st term of progression, d = common difference and n = number of terms. Either of the above two equations can be used to calculate the sum of an A.P.
 

Example
Find the sum of the terms in the progression 2, 5, 8, 11 … up to 15 terms.
Solution
From the given A.P., we can see that a = 2, d = 3 and n = 15.
 
The sum of n terms in an arithmetic progression is given by
 
Substituting the value of the terms in the above relation, we get
 
 

 

Example
Find the sum of the terms in the progression 5, 10, 15, 20, 25 … 100
Solution
In the above progression 100 is the 20th term (can be found using the relation to find the nth term).
Given that, n = 20
From the given progression, we can see that a = 5 and Tn = 100.
The sum of terms in the progression can be found by using the relation, 
Substituting the values, we get

 

Note: If the sum of terms in an A.P. is known and you are asked to find the terms, then the terms must be selected in the following ways, while solving problems:
When the sum of three terms is given, the terms must be selected as (a – d ), a, (a + d)
When the sum of four terms is given, the terms must be selected as (a – 3d ), (a – d ), (a + d ), (a + 3d )
When the sum of five terms is given, the terms must be selected as (a – 2d ), (a – d ), a, (a + d ), (a + 2d )

 

Example
Find the arithmetic progression, the sum of whose first 3 terms is 30 and the sum of their squares is 350.
Solution
We know the sum of 3 terms.
Hence, the terms shall be taken in the form (a – d), a, (a + d).
Given that, (a – d) + a + (a + d) = 30
⇒ 3a = 30 ⇒ a = 10
Sum of the squares of above 3 terms is,
(a – d)2 + a2 + (a + d)2 = 350
⇒ a2 + d 2 – 2ab + a2 + a2 + d 2 + 2ab = 350
⇒ 3a2 + 2d 2 = 350
Substituting the value of ‘a’, we get
⇒ 3(10)2 + 2d 2 = 350 ⇒ 2d 2 = 350 – 300
⇒ 2d 2 = 50 ⇒ d 2 = 25
⇒ d = ± 5
Using these values of a and d, we get the terms of the A.P. as follows:
(10 – 5), 10, (10 + 5) ... = 5, 10, 15 ...

 

Remember:

  • If we add or subtract a constant value to or from each term of an A.P., the resulting progression will also be an A.P.
  • If each term of an A.P. is multiplied or divided by any constant term, the resulting progression will also be an A.P.
  • If we add or subtract corresponding terms of two A.P.s, the resulting progression will also be an A.P.
  • If we multiply or divide corresponding terms of two A.Ps, the resulting progression will not be an A.P.

 

Example
If the 12th term of an A.P. is –13 and the sum of the first four terms is 24, what is the sum of the first 10 terms?
Solution
Given that T12 = –13, S4 = 24
 
 

 

Note:

  • Sum of 1st n odd numbers is
     
    S = 1 + 3 + 5 + ... + (2n – 1)
     
    S = n2
  • Sum of the 1st n natural numbers is
     
    S = 1 + 2 + 3 ... n
     
  • Sum of the squares of the 1st n natural numbers is
     
    S = 12 + 22 + 32 + ... n2
     
  • Sum of the cubes of the 1st n natural numbers is
     
    S = 13 + 23 + 33 + ... n3
     





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