Aptitude - Basic Arithmetic
Sequence
A sequence represents numbers formed in succession and arranged in a fixed order defined by a certain rule.
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.
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
