Arithmetic progression (a.p.), Mathematics

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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

                    Tn  = a + (n - 1)d

Generally the Tn  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 nth term? They are calculated as follows. We know that

                   T1     = a                = 3

                   T2     = a + d           = 3 + 2 = 5

                   T3     = a + 2d         = 3 + 2(2) = 7

                   T4     = a + 3d         = 3 + 3(2) = 9

                   T5     = a + 4d         = 3 + 4(2) = 11

                   :                                          :
                   :                                          :

           l = Tn        = a + (n - 1)d  = 3 + (n - 1)(2)

                                                = 3 + 2n - 2

                                                = 2n + 1


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