5.3 Arithmetic Series (1/5) In an arithmetic series each term increases by a constant amount (d) This means the difference between consecutive terms is.

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Presentation transcript:

5.3 Arithmetic Series (1/5) In an arithmetic series each term increases by a constant amount (d) This means the difference between consecutive terms is constant (d) Therefore: T11 – T10 = T37 – T36 = T317 – T316 = d

The terms of an arithmetic series are: + (a+1d) + (a+2d) + (a+3d) + … + [a+(n-1)d] + … Tn = a + (n – 1) d Value of nth term 1st term Common Difference Term number

Find the 101th term of the series 5.3 Arithmetic Series (3/5) Find the 101th term of the series 3 + 7+ 11 + … a = 3 n = 101 d = 11 – 7 = 4 Tn = a + (n – 1) d = 3 + (101 – 1) 4 = 403

What is the nth term of the series 5.3 Arithmetic Series (4/5) What is the nth term of the series 1 + 7+ 13 + … a = 1 d = 13 – 7 = 6 Tn = a + (n – 1) d = 1 + (n – 1) 6 = 1 + 6n – 6 Tn= 6n – 5

Which term of the series is 189 . 5.3 Arithmetic Series (5/5) Which term of the series is 189 . 9 + 16+ 21 + … Tn = a + (n – 1) d a = 9 189 = 9 + (n – 1) 5 Tn = 189 189 = 9 + 5n – 5 189 = 5n + 4 d = 16 – 9 5n = 185 =5 n = 36