Geometric progression(GP)
Geometric progression is a sequence of numbers in which the ratio between any two consecutive terms is constant. The constant ratio between the consecutive terms for the sequence is called the common ratio, and we can multiply it to any term of the sequence to find the next term.
For example 2, 4, 8, 16, 32, 64 is a geometric progression as the ratio between any two consecutive terms is constant, i.e., 64/32=32/16=16/8=8/4=4/2=2.
Important formulas for geometric progressions
Let, A be the first term, R be the common ratio, and N be the number of terms in the sequence.
 The nth^{ }term of the geometric progressions= A*R^{N1}

Sum of N terms of the geometric progressions
 If common ratio<1, Sum= A(1(R)^{N})/(1R)
 If common ratio<1, Sum= A((R)^{N}1)/(R1)
 For infinite geometric progression(common ratio<1), Sum=A/(1 R)

Geometric mean =
Harmonic progression(HP)
Harmonic progression is a sequence of numbers in which the multiplicative inverse or reciprocal of the terms forms an arithmetic progression.
For example 1, 3, 5, 7, 9, 11 is an arithmetic progression. Thus 1, 1⁄3, 1⁄5, 1⁄7, 1⁄9, 1⁄11 will be a harmonic progression. Inharmonic progression, the difference between reciprocals of any two consecutive terms is constant.
Important formulas for harmonic progressions
Let, A be the reciprocal of the first term of the harmonic progression, D be the common difference between the reciprocals of any two consecutive terms, and N be the number of terms in the sequence.
 N^{th }term of the harmonic progressions= 1/(A+(N1)D)
 Sum of N terms of the harmonic progressions= 2/N(2A+(N1)D)
 Harmonic mean =N/(1/A+1/(A+D)......+1/(last term))
Sample problems
Let’s see some problems to build a better understanding of these concepts
Problem1 Find the common difference of the following AP and write the AP up to 8 terms.
7, 12, 17, 22, 27...
Solution The common difference for an AP is the difference between any of its two consecutive terms. Thus for this AP, the common difference will be 127=5.
Since we are given the five terms of the AP, we can find the next terms by adding the common difference.
6^{th }term= 27+5=32
7^{th }term= 32+5=37
8^{th }term= 37+5=42
Thus the AP is 7, 12, 17, 22, 27, 32, 37, 42
Problem2 If a, b, c are in AP, find the value of a2b+c.
Solution Since a, b, c are in AP. Let's suppose the common difference of this AP is d.
Then the b=a+d, c=a+2d.
Now putting the value of b and c, in the expression a2b+c, it will become a2(a+d)+a+2d
=a2a2d+a+2d=0
Thus, the value of the expression a2b+c will be zero if a, b, c are in AP.
Problem3 The sum of the first six terms of an AP is 96, and the common difference is 4. Find the 4^{th }term of the AP.
Solution Let, the given AP has the first term as A, and it is given that total no. of terms=6, and the common difference is 4. Now, putting these values in the formula for the sum of an AP
1⁄2(N)(2*A +(N1)*D)= 6/2(2*A+5*4).
Since it is given that the sum of the AP is 96
=> 6/2(2*A+5*4)=96
=> A=6.
Thus, the required AP is 6, 10, 14, 18, 22. 26. The 4^{th }term of the AP is 18.
Problem4 Find the number of terms in the following GP and also find its sum.
2, 6, 18, …., 1458.
Solution To find the number of terms in the GP, we first need to find the common ratio of the progression. Since common ratio(R) = T_{n+1}/T_{n}=6/2=3.
Now, we also know that N^{th }term of a GP is A*R^{N1} where R is the common ratio.
Since the last term of the given GP is 1458, Now, equate it with the above formula to find the number of terms in the progression
=> 1458= A*R^{N1}
=> 2*729= 2*3^{N1}
=> 2*3^{6}= 2*3^{N1}
On comparing LHS with RHS
=> 6=N1
=> N=7
Problem5 If the sum of reciprocals of the first 13 terms of an HP is 260, find the 7^{th} term of the sequence.
Solution Sum of reciprocals of an HP refers to the sum of AP used to form the HP,
Let’ the first term of the AP be A, and common difference be D, then the sum of the AP will be 1⁄2(13)(2*A +(131)*D) Since the sum is given to be 260.
=> 1⁄2(13)(2*A +(131)*D)=260
=> 2*A +(131)*D=40
=> 2*A +12*D=40
=> A +6*D=20
Since the Nth term of an HP is 1/(A+(N1)D)
Thus, the 7th term of given HP will be 1/(A+6*D)=1/20.
Frequently Asked Questions
Q1. How can we check if the given sequence is an AP?
We can check if a sequence is an AP or not by comparing the consecutive terms of the sequence. If the difference between any two consecutive terms is constant, then the sequence is an arithmetic progression.
Q2. How to check if the given sequence is a GP?
We can check if a sequence is a GP or not by comparing the consecutive terms of the sequence. If the ratio between any two consecutive terms is constant, then the sequence is a geometric progression.
Q3. How to check if the given sequence is an HP?
We can check if a sequence is an HP or not by finding the reciprocals of its terms; if these reciprocals form an AP, the sequence is a harmonic progression.
Q4. What is the relation between Arithmetic mean, Geometric mean, and Harmonic mean?
The relation between Arithmetic mean Geometric mean, and Harmonic mean is as follows
Q5. Find the scenario in which all three values, i.e., Arithmetic mean, Geometric mean, and Harmonic mean for a sequence, can be equal?
Arithmetic, geometric, and Harmonic mean for a sequence can be equal when all the terms in the sequence are equal.
Key takeaways
This blog has discussed the important formulas and concepts regarding the progression, as it is one of the most important topics for aptitude exams.
We started with discussing the arithmetic progression, we learned the formulas to find the Nth term, the sum of the sequence up to the nth term, and the arithmetic mean of a sequence. Then we learned similar formulas for geometric and harmonic progressions as well. Lastly, we tried some problems to develop a better understanding of these concepts.
You can learn more about these concepts here. Visit Coding Ninjas Studio to practice similar questions.