A series is said to be in Arithmetic Progression (A.P.) if the consecutive numbers in the series differs by a constant value. This constant value is referred to as "common difference". The series in which the consecutive terms increases by a constant quantity, is referred to as an increasing series and if the terms decrease by a constant quantity it is referred to as a decreasing series. The series
3, 7, 11, 15, 19, .............
is an example of increasing series, while the one like
8, 2, -4, .........
is an example of decreasing series.
In an A.P. the first number is denoted by "a" and the common difference is denoted by "d". If we know the values of a and d, it is quite easy to get the terms of the Arithmetic Progression. In terms of a and d, the consecutive terms of arithmetic progression are
a, a + d, a + 2d, a + 3d, ......... a + nd
We observe that the first term is a, the second term is a + d, the third term being a + 2d. The point to note is that for the first term the coefficient of d is zero, for the second term it is one and for the third term it is 2. By observing this pattern can we conclude that the coefficient of nth term is n - 1? Yes, we can. In fact, the nth term is given by
T_{n} = a + (n - 1)d
Generally the T_{n} which is the last term is also denoted by "l" (small alphabet 'l'). That is, l = a + (n - 1)d.
Now let us look at an example.
Example
If the first term of an A.P. 'a' = 3 and the common difference 'd' = 2, what are the first five terms of the series and what would be the n^{th} term? They are calculated as follows. We know that
T_{1} = a = 3
T_{2} = a + d = 3 + 2 = 5
T_{3} = a + 2d = 3 + 2(2) = 7
T_{4} = a + 3d = 3 + 3(2) = 9
T_{5} = a + 4d = 3 + 4(2) = 11
: :
: :
l = T_{n} = a + (n - 1)d = 3 + (n - 1)(2)
= 3 + 2n - 2
= 2n + 1