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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

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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

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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

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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)

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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!!

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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

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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

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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)

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