Find the 21" term of AP: 21, 29, 37, 45, ....
Answers
Find the difference between the members
a2-a1=29-21=8
a3-a2=37-29=8
a4-a3=45-37=8
The difference between every two adjacent members of the series is constant and equal to 8
General Form: an=a1+(n-1)d
an=21+(n-1)8
a1=21 (this is the 1st member)
an=45 (this is the last/nth member)
d=8 (this is the difference between consecutive members)
n=4 (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:
21+29+37+45
This sum can be found quickly by taking the number n of terms being added (here 4), multiplying by the sum of the first and last number in the progression (here 21 + 45 = 66), and dividing by 2:
n(a1+an)
2
4(21+45)
2
The sum of the 4 members of this series is 132
This series corresponds to the following straight line y=8x+21
Finding the nth element
a1 =a1+(n-1)*d =21+(1-1)*8 =21
a2 =a1+(n-1)*d =21+(2-1)*8 =29
a3 =a1+(n-1)*d =21+(3-1)*8 =37
a4 =a1+(n-1)*d =21+(4-1)*8 =45
a5 =a1+(n-1)*d =21+(5-1)*8 =53
a6 =a1+(n-1)*d =21+(6-1)*8 =61
a7 =a1+(n-1)*d =21+(7-1)*8 =69
a8 =a1+(n-1)*d =21+(8-1)*8 =77
a9 =a1+(n-1)*d =21+(9-1)*8 =85
a10 =a1+(n-1)*d =21+(10-1)*8 =93
a11 =a1+(n-1)*d =21+(11-1)*8 =101
a12 =a1+(n-1)*d =21+(12-1)*8 =109
a13 =a1+(n-1)*d =21+(13-1)*8 =117
a14 =a1+(n-1)*d =21+(14-1)*8 =125
a15 =a1+(n-1)*d =21+(15-1)*8 =133
a16 =a1+(n-1)*d =21+(16-1)*8 =141
a17 =a1+(n-1)*d =21+(17-1)*8 =149
a18 =a1+(n-1)*d =21+(18-1)*8 =157
a19 =a1+(n-1)*d =21+(19-1)*8 =165
a20 =a1+(n-1)*d =21+(20-1)*8 =173
a21 =a1+(n-1)*d =21+(21-1)*8 =181
a22 =a1+(n-1)*d =21+(22-1)*8 =189
a23 =a1+(n-1)*d =21+(23-1)*8 =197
a24 =a1+(n-1)*d =21+(24-1)*8 =205
a25 =a1+(n-1)*d =21+(25-1)*8 =213
a26 =a1+(n-1)*d =21+(26-1)*8 =221
a27 =a1+(n-1)*d =21+(27-1)*8 =229
a28 =a1+(n-1)*d =21+(28-1)*8 =237
a29 =a1+(n-1)*d =21+(29-1)*8 =245