A progression is an ordered list of numbers in which each term is related to the previous one by a definite rule. This rule determines how the sequence grows or changes, making it predictable and structured.
Common Types of Progressions:
Arithmetic Progression (AP)
A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series. This fixed number is called the common difference.
For example: 2, 4, 6, 8, 10 is an AP because the difference between any two consecutive terms in the series (common difference) is the same (4 – 2 = 6 – 4 = 8 – 6 = 10 – 8 = 2).
nth term of an AP = a + (n-1) d
where,
- a → First term of the arithmetic progression
- n → Position (term number) that we want to find
- d → Common difference between consecutive terms
Arithmetic Mean = Sum of all terms in the AP / Number of terms in the AP
Sum of ‘n’ terms of an AP = 0.5 n (first term + last term) = 0.5 n [ 2a + (n-1) d ]
Geometric Progression (GP)
A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by multiplying a fixed number by the previous number in the series. This fixed number is called the common ratio.
For example: 2, 4, 8, 16 is a GP because the ratio of any two consecutive terms in the series is the same (4 / 2 = 8 / 4 = 16 / 8 = 2).
nth term of a GP = a rn-1
where,
- a → First term of the geometric progression
- r → Common ratio (the number by which each term is multiplied to get the next term)
- n → Position or term number that we want to find
Geometric Mean = nth root of product of n terms in the GP
Sum of ‘n’ terms of a GP (r < 1) = [a (1 – rn)] / [1 – r]
Sum of ‘n’ terms of a GP (r > 1) = [a (rn – 1)] / [r – 1]
Sum of infinite terms of a GP (r < 1) = (a) / (1 – r)
Harmonic Progression (HP)
A sequence of numbers is called a harmonic progression if the reciprocals of the terms are in AP.
In simple terms, a, b, c, d, e, f are in HP if 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP.
- Harmonic Mean = (2 a b) / (a + b)
For two numbers, if A, G, and H are respectively the arithmetic, geometric, and harmonic means, then.
- A ≥ G ≥ H
- A H = G2, i.e., A, G, H are in GP
Solved Questions and Answers
Question 1 : Find the nth term for the arithmetic progression (AP): 5, 9, 13, 17, …
Solution:
Here,
a = 5,
difference (d) = 9 − 5 = 4We know that the nth term of an AP is given by: nth term = a + (n − 1)d
⇒ nth term for the given AP: = 5 + (n − 1)⋅4
⇒ nth term for the given AP: = 5 + 4n − 4
⇒ nth term for the given AP: = 4n + 1
We can verify the answer by putting values of ‘n’.
⇒ n = 1 → First term = 4⋅1 + 1 = 5
⇒ n = 2 → First term = 4⋅2 + 1 = 9
⇒ n = 3 → First term = 4⋅3 + 1 = 13
⇒ n = 4 → First term = 4⋅4 + 1 = 17
Question 2 : The sum of three sequential numbers of a GP is 28, and their product is 512. Find the numbers.
Solution:
Let the numbers be a/r, a, ar.
a/r + a + ar = 28
This simplifies to: a(1 + r + r2)/r = 28
Also, it is given that the product is 512: (a/r) × a × (ar) = 512
⟹ a3 = 512
⟹ a = 8
Now substituting a back into the sum equation: 8(1 + r + r2) / r = 28
⟹ 1 + r + r2/ r = 28/8 = 7/2
Multiplying through by r: 1 + r + r2 = 7/2r
⟹ 2 + 2r + 2r2 = 7r
Rearranging gives: 2r2 − 5r + 2 = 0
Factoring the quadratic: (2r − 1)(r − 2) = 0
Thus, r = 1/2 or r = 2.
The required numbers are 4,8 and 16.
Question 3 : For the elements 4 and 6, verify that A ≥ G ≥ H.
Solution :
A = Arithmetic Mean = (4 + 6) / 2 = 5
G = Geometric Mean =
\sqrt{{4}\times{6}} = 4.8989H = Harmonic Mean = (2 x 4 x 6) / (4 + 6) = 48 / 10 = 4.8
Therefore, A ≥ G ≥ H
Question 4 : Find the sum of the infinite series 32, 16, 8, 4, ....
Solution :
First term, a = 32
Common ratio, r = 16 / 32 = 8 / 16 = 4 / 8 = 1 / 2 = 0.5
We know that for an infinite GP, Sum of terms = a / (1 - r)
=> Sum of terms of the GP = 32 / (1 - 0.5) = 32 / 0.5 = 64
Question 5 : The sum of three numbers in a GP is 26 and their product is 216. find the numbers.
Solution :
Let the numbers be a/r, a, ar.
=> (a / r) + a + a r = 26
=> a (1 + r + r2) / r = 26
Also, it is given that product = 216
=> (a / r) x (a) x (a r) = 216
=> a3 = 216
=> a = 6
=> 6 (1 + r + r2) / r = 26
=> (1 + r + r2) / r = 26 / 6 = 13 / 3
=> 3 + 3 r + 3 r2 = 13 r
=> 3 r2 - 10 r + 3 = 0
=> (r - 3) (r - (1 / 3) ) = 0
=> r = 3 or r = 1 / 3
Thus, the required numbers are 2, 6 and 18.