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 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
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I joined CATKing classes in April 2022 for CAT 2022 and other OMETs. The classes were held daily and there were special fast paced quants classes on weekends. Sudanshu Sir is amazing and so is Rahul Sir. Both are experts in their fields. The Alumini connect sessions which are held with former CATKing students also help you plan your studies accordingly. The course material is divided into different LODs – Level of Difficulties, which you can access based on your level of preparation. Finally, the best thing about CATKing is the mocks. Their mocks are the closest to the difficulty level of the actual CAT papers. I got 77 in the final mock before CAT and got 74 raw score in CAT 22. These were my final scores after studying 5 months at CATKing: CAT 22 – 97.98 IIFT 22 – 99.59 SNAP 22 – 99.31 NMAT – 256 The GDPI prep for the interviews for your calls is all great. you have college seniors helping you out taking your mock interviews. That was very helpful as well. You get compiled dossiers for each college to help you prepare. They helped me convert some of my best calls – NITIE, IIFT, NMIMS, SCMHRD, All CAP IIMs. Finally I would be joining IIFT Delhi, all thanks to CATKing.
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Hello, Though I am not the right authority to decide for anyone as to how they can prepare best for CAT, here’s my take on how CATKing helped me . I scored 97.2%ile in CAT 2022, and I am joining IIM Raipur for PGP in Management 2023-25. As for my prepartion stratergy, it was non-existent if I consider my peers and all their efforts. With no disrespect to anyone, I got really lucky with my score. I was not serious with my preparation and was not looking forward to joining any colleges this year, and hence I was completely clueless about MBA and its proceedings. I had only applied for 11 colleges through CAT application and IMT. So even my options were very short. After the result declarations I was at my wits end and started my MBA preparations journey, where the first person I came across was Sumit Sir. His tips and tricks on cracking Interviews enticed me and following his suggestions via DM, I joined the WAT PI batch. It was monumental in my preparation for all the interviews ( I got PI calls for all except BLACK). Their sessions and dock is extensive and the community is very enriching and supportive. I converted all my calls and I am at the moment waitlisted for Indore, not a single rejection. If you are looking for proper guidance, support and have faith in yourself, CATKing delivers what they promise. Hope this helps you make your decision.
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