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Arithmetic series

 S n =. [2a + (n – 1)d].  S n =. [a + L]. Arithmetic series. The sum of all the terms of a arithmetic progression is called a arithmetic series. General formula for the sum of a AP. Sum of the first n terms:. S n = a + [a + d] + [a + 2d] + [a + (n - 1)d] [1].

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Arithmetic series

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  1.  Sn = [2a + (n – 1)d]  Sn = [a + L] Arithmetic series The sum of all the terms of a arithmetic progression is called a arithmeticseries. General formula for the sum of a AP Sum of the first n terms: Sn = a + [a + d] + [a + 2d] + [a + (n - 1)d] [1] Write the sum in reverse order: Sn = [a +(n – 1)d] + . . [a + 2d] + [a + d]+ a [2] Add [1] and [2] 2Sn = All the terms are 2a + (n- 1)d and there are n of them. 2Sn = n[ 2a + (n – 1)d] If you know the last term L:

  2.  Sn = [a + L] The sum of the first n natural numbers The first term is a = 1 The last term is L = n There are n number of terms a = 1, L = n and n = n Example: find the sum of first 49 numbers a = 1, L = 49 and n = 49 = 49 x 25 = = 1225

  3.  Sn = [2a + (n – 1)d] Examples In an arithmetic series, the 3rd term is 17 and the 10th term is 52. Find the 1st term, the common difference and the sum of the first 20 terms. nth term = a + d(n- 1) 3rd term : a + 2d = 17 [1] 10th term : a + 9d = 52 [2] [2] – [1]: 7d = 35  d = 5 The common ratio [1]: a + 2 x 5 = 17  a = 7 The 1st term Formula for the sum Sn = 10[ 2 x 7 + 19 x 5 ] = 1090

  4.  Sn = [2a + (n – 1)d] Examples In an AP, S8 = 56 and S20 = 260. Calculate a the first term and the common difference d. 4[ 2a + 7d ] = 56: 2a + 7d = 14 [1] 10[ 2a + 19d ] = 260: 2a + 19d = 26 [2] [2] – [1]: 12d = 12  d = 1 The common ratio [1]: 2a + 7 x 1 = 14  a = 3.5 The 1st term The first term is 3.5 and the common difference is 1.

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