1. Arithmetic Series Understand the difference between a sequence and a series Proving the nth term rule Proving the formula to find the sum of an arithmetic series

2. Consider the infinite sequence 4,7,10,13,…. If the terms of the sequence are added this becomes a finite series 4+7+10+13 In an arithmetic series the difference between the terms is constant. The difference is called the common difference

3. An arithmetic series is also known as an arithmetic progression (AP) Using the sequence 4, 7, 10, 13… a=1 st term of the sequence d=common difference n1234 471013 3n+1 aa+da+2da+3d So the nth term would be…. a + (n-1)d

4. Proof the the sum of an Arithmetic Series n1234…..1920 3n+1471013….. 6158 Call the sum of the terms S n S n = 4 + 7 + 10 + 13 + ….. + 58 + 61 S n = 61+58 + 55+ 52 + ….. + 4 + 7 Reverse the order Add the two series together 2S n = 65 + 65 + 65 + 65 + ….. + 65 + 65 2S n = 65x 20 (because there are 20 terms) 2S n = 1300 S n = 650 (divide by 2)

5. Proof the the sum of an Arithmetic Series n1234…..n-1n aa+da+2da+3d…..L-dL S n = a + (a+d) + (a+2d) + (a+3d) + ….. + (L-2d) + (L-d) + L S n = L + (L-d) + (L-2d) + (L-3d) + ….. +(a+2d) + (a+d)+ a Sum the first n terms then reverse the order Add the two series together 2S n = (a+L)+(a+L)+ (a+L) + (a+L) + ….. + (a+L) + (a+L)+(a+L) 2S n = n(a+L) (because there are n terms) S n = n(a+L) 2 a=first term, d=common difference, L=last term Nearly there!!

6. Proof the the sum of an Arithmetic Series L (the last term) is also the nth term which we know has the formula a+(n-1)d so if we substitute for L in the formula above we get…. a=first term, d=common difference, L=last term S n = n(a+L) 2 S n = n[a+a+(n-1)d] 2 S n = n[2a+(n-1)d] 2 You need to learn this formula

7. EXAMPLE 1 Find the sum of the first 30 terms in the series 3+9+15+… a=3, d=6, n=30 Using the formula S n = n[2a+(n-1)d] 2 S n = 30[2x3+(30-1)6] 2 S n = 15[6+(29x6)] S n = 15x180 = 2700

8. EXAMPLE 2 a)Find the nth term of the arithmetic series 7+11+15+.. b)Which term of the sequence is equal to 51? c)Hence find 7+11+15+…+51 a) a=7, d=4 so the nth term is 4n+3 c) Using the formula S n = n[2a+(n-1)d] a=7, d=4 and n=12 2 S n = 12[2x7+(12-1)4] 2 S n = 6[14+(11x4)] S n = 6x58 = 348 b) 4n+3= 51 4n = 48 (subtract 3) n = 12 (divide by 4)