Arithmetic progression - SS2 Mathematics Lesson Note

A set of numbers arranged consecutively is called a sequence. A sequence is an arithmetic sequence or arithmetic progression or linear progression if and only if the difference any member and the member immediately after it is constant. The members of these sequence are called terms of the sequence and they sum up to a series.

In an A.P., where \(a\) is the first term, \(d\) is the difference between consecutive terms (the common difference), then \(T_{n}\), the \(n\)th term or the general term of the sequence is gotten by

\[T_{n} = a + (n - 1)d\]

The sum of \(n\) terms in an A.P. is the series:

\(S_{n} = \frac{n}{2}\lbrack 2a + (n - 1)d\rbrack\) OR

\({\sum_{1}^{n}{S_{n} =}S}_{n} = \frac{n}{2}\left\lbrack a + T_{n} \right\rbrack\)(\(a\) is the 1^st term and \(T_{n}\) is the last term).

Example 1 Find the 9^th term in the sequence \(6,\ 11,\ 16,\ 21,\ 26,\ \ldots\) and find the sum of the first \(10\) terms of the series.

Solution

The 9^th term of the Arithmetic progression \(T_{n} = a + (n - 1)d\)

Where \(n\ = \ 9,\ a\ = \ 6\ \) and \(d\ = \ 11\ –\ 6\ = \ 5\)

\[T_{9} = 6 + (9 - 1)5\]

\[T_{9} = 6 + (8)5\]

\[T_{9} = 6 + 40\]

\[T_{9} = 46\]

\[S_{n} = \frac{n}{2}\lbrack 2a + (n - 1)d\rbrack\]

\[S_{10} = \frac{10}{2}\lbrack 2(6) + (10 - 1)5\rbrack\]

\[S_{10} = 5\lbrack 12 + (9)5\rbrack\]

\[S_{10} = 5\lbrack 12 + 45\rbrack\]

\[S_{10} = 5\lbrack 57\rbrack\]

\[S_{10} = 285\]

Example 2 Find the first three terms given by the \(n\)th term, \(\frac{n}{n + 2}\)

Solution Given \(T_{n} = \frac{n}{n + 2}\)

The first three terms of the A.P. \(T_{1} = \frac{1}{1 + 2} = \ \frac{1}{3}\); \(T_{2} = \frac{2}{2 + 2} = \ \frac{2}{4} = \ \frac{1}{2}\); \(T_{3} = \frac{3}{3 + 2} = \ \frac{3}{5}\)

The mean of an A.P. is usually the middle term of the A.P, calculated as \(\frac{S_{n}}{n}\) or \(S_{n} = \frac{1}{2}\lbrack a + T_{n}\rbrack\). For instance, the mean of the first 10 terms in example 1 is \(\frac{S_{n}}{n} = \ \frac{285}{10} = 28.5\)