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 2, 5, 8, 11, 14, 17. You can notice that in this example the difference between any two terms of the sequence is 3 (5-2=3, 8-5=3, 11-8=3 and so on.) Therefore, in this AP the common difference ‘d’ is 3 and the first term ‘a’ is 2.

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence. For example, the series of natural numbers: 1, 2, 3, 4, 5, 6,… is an Arithmetic Progression, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). Even in the case of odd numbers and even numbers, we can see the common difference between two successive terms will be equal to 2.

Example 1

The first term of A.P. is 1 and the common difference is 5 calculates the 7th term of this sequence.

Solution

The sequence can be formed as 1, 6, 11, 16… and we are interested to get the 7th term of this sequence. We can write it in the form a + 6d and thus get 31. Generalizing it, we get a + (n1)d. Hence, the formula we get for the nth term of A.P. is denoted by

Tn = a+ (n-1)d.

What if you have to calculate the sum to n in terms of the above sequence then you can use the formula given as

Sn = n/2 {2a + (n-1) d} or Sn = n/2 {a + l},

where l is the last term of A.P.
Thus, using this formula we get S7 = 112

Example 2

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

Solution

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

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

Arithmetic 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 the 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 1,3,5,7,9,11,…. You can notice that in this example the difference between any two terms of the sequence is 2 Therefore, in this AP the common difference ‘d’ is 2 and the first term ‘a’ is 1.

Tn = a+ (n-1)d.

Question 3
What is the 25th term of the A.P. 3,7,11,15,19….

Solution
Tn = a+ (n-1)d.
T25= 3+ 24*4
= 99
Answer 99

Question 4
Find the common difference for the following AP: 10, 20, 30, 40, 50.

Solution:
Given AP: 10, 20, 30, 40, 50
Common difference:
d = 20 – 10 = 10
d = 30 – 20 = 10
d = 40 – 30 = 10
d = 50 – 40 = 10.
Hence, the common difference for the sequence, 10, 20, 30, 40, 50 is 10.

Question 5
Is a, 2a, 3a, 4a, … an arithmetic progression?

Solution:
Given sequence: a, 2a, 3a, 4a, …
To check whether the given sequence is AP or not, we have to find the common difference.
Hence, d = 2a – a = a
d = 3a – 2a = a
d = 4a – 3a = a
Hence, the common difference is “a”.
Therefore, the sequence a, 2a, 3a, 4a,… is an arithmetic progression.

Question 6
The sequence 28, 22, x, y, 4 is an AP. Find the values of x and y.

Solution:
Given AP: 28, 22, x, y, 4
Here, first term, a = 28
Common difference, d = 22 – 28 = -6
Hence, x = 22 – 6 = 16
y = 16 – 6 = 10.
Hence, the values of x and y are 16 and 10, respectively.

Question 7
Find the 5th term of the arithmetic progression 1, 4, 7, ….

Solution:
Given AP: 1, 4, 7, …
a = 1
d = 4 – 1 = 3
n = 5
As we know,
The nth term of AP = a + (n-1)d
Hence, 5th term of AP = 1 + (5-1)3
= 1+(4)3
= 1 + 12
= 13.
Hence, the 5th term of AP is 13.

Question 8
Find the 17th term of AP 4, 9, 14, …

Solution:
Given AP: 4, 9, 14, …
Here, a = 4
d = 9 – 4 = 5
n = 17.
Now, substitute the values in the formula a+(n-1)d,
17th term of AP = 4+(17-1)5
= 4+16(5)
= 4+80 = 84
Hence, the 17th term of AP is 84.

Question 9
If the first, second, and last terms of the AP are 5, 9, and 101, respectively, find the total number of terms in the AP.

Solution:
Given: First term, a = 5
Common difference, d = 9 – 5 = 4
Last term, an = 101
Now, we have to find the value of “n”.
Hence, an = a+(n-1)d
Substituting the values, we get
5+(n-1)4 = 101
5+4n-4 = 101
4n+1= 101
4n = 100
n=100/4 = 25
Hence, the number of terms in the AP is 25.

Question 10
Which term of AP 27, 24, 21, … is 0?

Solution:
Given AP: 27, 24, 21, …
Here, a = 27
d = 24 – 27 = -3.
Also given that an = 0
Now, we have to find the value of n.
Hence, an = a+(n-1)d
0 = 27 +(n-1)(-3)
0 = 27 -3n +3
0 = 30 – 3n
3n = 30
Hence, n = 10
Therefore, the 10th term of AP is 0.

Question 11
What is the sum till the 25th term of the A.P. 3,7,11,15,19….

Solution
Sn = n/2 {2a + (n-1) d}
= 25/2{ 2* 3 + 24*4}
=1275
Answer 1275