1.Sequence
A sequence represents numbers formed in succession and arranged in a fixed order defined by a certain rule.
2.Airthmetic Progression ( A.P.)
It is a type of sequence where each number/term(except first term) differs from its preceding number by a constant. This constant is termed as common difference.
3.A.P. Terminologies
- First number is denoted as ‘a’.
- Common difference is denoted as ‘d’.
- nth number is denoted as ‘Tn‘.
- Sum of n number is denoted as ‘Sn‘.
A.P. Examples
- 1, 3, 5, 7, … is an A.P. where a = 1 and d = 3 – 1 = 2.
- 7, 5, 3, 1, – 1 … is an A.P. where a = 7 and d = 5 – 7 = -2.
General term of A.P.
Tn = a + (n - 1)d
Where a is first term, n is count of terms and d is the difference between two terms.
Sum of n terms of A.P.
Sn = (n/2)[2a + (n - 1)d
Where a is first term, n is count of terms and d is the difference between two terms. There is another variation of the same formula:
Sn = (n/2)(a + l)
Where a is first term, n is count of terms, l is the last term.
Geometrical Progression, G.P.
It is a type of sequence where each number/term(except first term) bears a constant ratio from its preceding number. This constant is termed as common ratio.
G.P. Terminogies
- First number is denoted as ‘a’.
- Common ratio is denoted as ‘r’.
- nth number is denoted as ‘Tn‘.
- Sum of n number is denoted as ‘Sn‘.
G.P. Examples
- 3, 9, 27, 81, … is a G.P. where a = 3 and r = 9 / 3 = 3.
- 81, 27, 9, 3, 1 … is a G.P. where a = 81 and r = 27 / 81 = (1/3).
General term of G.P.
Tn = ar(n-1)
Where a is first term, n is count of terms, r is the common ratio
Sum of n terms of G.P.
Sn = a(1 - rn)/(1 - r)
Where a is first term, n is count of terms, r is the common ratio and r < 1. There is another variation of the same formula:
Sn = a(rn - 1)/(r - 1)
Where a is first term, n is count of terms, r is the common ratio and r > 1.
Arithmetic Mean
Airthmetic mean of two numbers a and b is:
Arithmetic Mean = (1/2)(a + b)
Geometric Mean
Geometric mean of two numbers a and b is
Geometric Mean = √ab
General Formulaes
1 + 2 + 3 + ... + n = (1/2)n(n+1)
12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6
13 + 23 + 33 + ... + n3 = [(1/2)n(n+1)]2
Very good brother…. Must useful topic
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Very good brother
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