Progression in maths can be defined as sequence of terms that increases in a particular pattern. There are different types of progression in Math which are mentioned below:
Geometric ProgressionBack to Top
The nth term of geometric progression is represented as arn-1. And for finding nth term of a series in G.P. we need to be aware about its first term. Recursive relation can also be followed by this kind of G.P. series given as an is equals to (r * an - 1). Common ratio is not specified as to be positive or negative; it can be anything positive as well as negative. Sum of 'n' terms in a geometric progression series is represented as follows:
Sn = a (1 – rn+1) / (1 - r) when series starts from initial value and for any middle element it is represented as:
Sn = a (rm – rn+1) / (1 - r) where 'm' is the degree of initial term when it is not zero.
For infinite series in geometric progressions, sum of series is represented as
Sn = a / (1 - r).
By Arithmetic Progression, we Mean that it is a sequence in which each term except first term differs from its preceding term by a constant.
Constant difference is called as common difference of Arithmetic Progressions. To Define Arithmetic Progression, we say that it is the series of numbers which has first term as “a”, common difference “ d” and we define n th term by an. Arithmetic Progression Definition also tells us that n th term of an A.P. with first term 'a' and common difference 'd' is given as follows:
an = a + (n – 1) * d,
Arithmetic Progression Problems solving is always convenient to make different choices based on following :
i) 3 numbers in AP as (a – d), a, (a + d),
ii) 4 numbers in AP as (a – 3d), (a – d), (a + d), (a + 3d),
iii) 5 numbers in AP as (a – 2d), (a – d), a , (a + d), (a + 2d),
iv) 6 numbers in AP as ((a – 5d), (a – 3d), (a – d), (a + d), (a + 3d), (a + 5d).
Arithmetic Progression Proof of two theorems are as follows:
Show that the n th term of an AP with first term 'a' and common difference 'd' is given by:
an = a + (n – 1)* d
Proof : Let us consider an AP with first term 'a' and common difference 'd'. Then, AP is given by:
a, a + d, a + 2d, a + 3d, …….
In this AP we have:
First term, a1 = a = a + 0 * d = a + (1 – 1) * d,
Second term, a2 = (a + d) = a + (2 - 1) * d,
Third term, a3 = (a + 2d) = a + (3 – 1) * d,
Thus we say that,
Nth term an = a + (n – 1) d,
So we say that an = a + (n – 1) d.
To show that n th term from end of an AP, with first term 'a', common difference 'd' and last term 'l' is given by [l – (n - 1) d],
Proof: If 'a' is first term, 'd' is common difference and 'l' is last term of given AP
Then the AP is given by: a, a + d, a + 2d, . . . . . , (l – 2d), (l – d), l.
So we say that last term will be = l = l – (1 - 1) d,
Second last term = (l – d) = l – (2 - 1) d,
Third last term = (l – 2d) = l – (3 – 1),
So nth last term will be = l – (n – 1)d.
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Arithmetic Progression FormulaBack to Top
Here we will focus on Arithmetic Progression.
A sequence of number in which difference between two successive Numbers is a constant is known as arithmetic progression. Representation of arithmetic progression is given as i, i + d, i + 2d, i + 3d, i + 4d …...... i + (n – 1) d, here the value of 'i' represents the initial value and 'd' represents the common difference. For example: 1, 4, 6, 8, 10, 12 here initial value of A.P is 1 and difference between two numbers is 2. Formula for arithmetic progression is given below.
When we want n th term of A.P then formula to find n th term is given by = i + (n – 1) d.
When we want sum of 'n' term in A.P then formula to find n th term is given by = n / 2 (2i + (n – 1) d).
We can find sum of Square of numbers in an A.P then formula to find sum of square of numbers is given by: n (n + 1) (2n + 6) / 6.
If p, q, r are given in A.P then value of 'b' is given as: b = (a + c) / 2. This is all about arithmetic progression formulas. It will be more clear with help of an example:
For example: 'n th' term of a sequence is given as 11, initial term is '1' and common difference is given as '2', which term is that?
Solution: As we know that n th term of sequence is given as: a + (n – 1) d.
Here initial value i = 1, common difference = 2 and n th term sn = 11. We get:
⇒ 11 = 1 + (n – 1) 2,
⇒ 11 – 1 = (n – 1) 2,
⇒ 10/2 = n – 1,
⇒ 5 + 1 = n.
So 6th term of sequence is 11.
Harmonic ProgressionBack to Top
Tn = 1 / a + (n - 1) * d,
Where Tn is general term of harmonic progression. 'a' is named as first term of concordant arithmetic progression 'd' is common difference and 'n' is number of terms.
There are some important points which will help to solve harmonic progression problems–
1. To solve harmonic progression problems, all terms of harmonic progression first should be written in Arithmetic progression then need to apply all the formulae of A.P. This is the first process to solve any problem of harmonic progressions.
2. Likewise, there is no formula to calculate sum of 'n' terms of harmonic progression so again apply above method means formula of sum of 'n' terms in A.P.
Now general formula of harmonic progression is its harmonic Mean. If there are two entities 'a' and 'b' and 'h' is harmonic mean between these entities, then 'h' can be calculated with help of following formula:
h = 2ab / (a + b),
So, in this case a, h, b will be in harmonic progression and 1 / a, 1 / h and 1 / b will be in arithmetic progression.
Formula to calculate 'n' harmonic means between two quantities 'a' and 'b' can be evaluated as follows–
Suppose that h1, h2, h3… hn are 'n' harmonic means then 1 / a, 1 / h1, 1 / h2 … 1 / hn, 1 / b will be in arithmetic progression and common difference for this A.P. is given by:
d = a – b / a b (n + 1)
So 'n' arithmetic means can be written between 'a' and 'b' as (1 / a) + d, (1 / a) + 2d, (1 / a) + 3d … (1 / a) + n d which can be simplified as (1 + ad) / a, (a + 2ad) / a, (1 + 3ad) / a … (1 + n a d) / a.
So n harmonic means between a and b will be a / (1 + ad), a / (1 + 2ad), a / (1 + 3ad) … a / (1 + n a d).
The relation between H.P., A.P. and G.P between two entities 'a' and 'b' is given as:
G2 = A * H,
Where 'A' is Arithmetic Mean, 'G' is geometric mean and 'H' is harmonic mean and one more Point to be noted is that A, G, H always lie in following order.
A > G > H
That means arithmetic mean is greatest in all means and harmonic mean is least one and basic concept which can be noted in whole article is the definition of harmonic progression which says that harmonic progressions are always reciprocal of terms of arithmetic progression.