a2-a1=18-9=9

a3-a2=27-18=9

a4-a3=36-27=9

a5-a4=45-36=9

The difference between EVERY two ADJACENT members of the SERIES is constant and equal to 9

General Form: an=a1+(n-1)d

an=9+(n-1)9

a1=9 (this is the 1ST member)

an=45 (this is the last/nth member)

d=9 (this is the difference between consecutive members)

n=5 (this is the number of members)

Sum of finite series members

The sum of the members of a finite arithmetic progression is called an arithmetic series.

Using our example, consider the sum:

9+18+27+36+45

This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 9 + 45 = 54), and DIVIDING by 2:

n(a1+an)

2

5(9+45)

2

The sum of the 5 members of this series is 135

This series corresponds to the following straight line y=9x+9

Finding the nth element

a1 =a1+(n-1)*d =9+(1-1)*9 =9

a2 =a1+(n-1)*d =9+(2-1)*9 =18

a3 =a1+(n-1)*d =9+(3-1)*9 =27

a4 =a1+(n-1)*d =9+(4-1)*9 =36

a5 =a1+(n-1)*d =9+(5-1)*9 =45

a6 =a1+(n-1)*d =9+(6-1)*9 =54

a7 =a1+(n-1)*d =9+(7-1)*9 =63

a8 =a1+(n-1)*d =9+(8-1)*9 =72

a9 =a1+(n-1)*d =9+(9-1)*9 =81

a10 =a1+(n-1)*d =9+(10-1)*9 =90

a11 =a1+(n-1)*d =9+(11-1)*9 =99

a12 =a1+(n-1)*d =9+(12-1)*9 =108

a13 =a1+(n-1)*d =9+(13-1)*9 =117

a14 =a1+(n-1)*d =9+(14-1)*9 =126

a15 =a1+(n-1)*d =9+(15-1)*9 =135

a16 =a1+(n-1)*d =9+(16-1)*9 =144

a17 =a1+(n-1)*d =9+(17-1)*9 =153

a18 =a1+(n-1)*d =9+(18-1)*9 =162

a19 =a1+(n-1)*d =9+(19-1)*9 =171

a20 =a1+(n-1)*d =9+(20-1)*9 =180

a21 =a1+(n-1)*d =9+(21-1)*9 =189

a22 =a1+(n-1)*d =9+(22-1)*9 =198

a23 =a1+(n-1)*d =9+(23-1)*9 =207

a24 =a1+(n-1)*d =9+(24-1)*9 =216

a25 =a1+(n-1)*d =9+(25-1)*9 =225

a26 =a1+(n-1)*d =9+(26-1)*9 =234

a27 =a1+(n-1)*d =9+(27-1)*9 =243

a28 =a1+(n-1)*d =9+(28-1)*9 =252

a29 =a1+(n-1)*d =9+(29-1)*9 =261

a30 =a1+(n-1)*d =9+(30-1)*9 =270