complete the number pattern 13,23,33,43
Answers
Answer:
Your input 13,23,33,43,53,63,73,83 appears to be an arithmetic sequence
Find the difference between the members
a2-a1=23-13=10
a3-a2=33-23=10
a4-a3=43-33=10
a5-a4=53-43=10
a6-a5=63-53=10
a7-a6=73-63=10
a8-a7=83-73=10
The difference between every two adjacent members of the series is constant and equal to 10
General Form: a
n
=a
1
+(n-1)d
a
n
=13+(n-1)10
a1=13 (this is the 1st member)
an=83 (this is the last/nth member)
d=10 (this is the difference between consecutive members)
n=8 (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:
13+23+33+43+53+63+73+83
This sum can be found quickly by taking the number n of terms being added (here 8), multiplying by the sum of the first and last number in the progression (here 13 + 83 = 96), and dividing by 2:
n(a1+an)
2
8(13+83)
2
The sum of the 8 members of this series is 384
This series corresponds to the following straight line y=10x+13
Finding the n
th
element
a1 =a1+(n-1)*d =13+(1-1)*10 =13
a2 =a1+(n-1)*d =13+(2-1)*10 =23
a3 =a1+(n-1)*d =13+(3-1)*10 =33
a4 =a1+(n-1)*d =13+(4-1)*10 =43
a5 =a1+(n-1)*d =13+(5-1)*10 =53
a6 =a1+(n-1)*d =13+(6-1)*10 =63
a7 =a1+(n-1)*d =13+(7-1)*10 =73
a8 =a1+(n-1)*d =13+(8-1)*10 =83
a9 =a1+(n-1)*d =13+(9-1)*10 =93
a10 =a1+(n-1)*d =13+(10-1)*10 =103
a11 =a1+(n-1)*d =13+(11-1)*10 =113
a12 =a1+(n-1)*d =13+(12-1)*10 =123
a13 =a1+(n-1)*d =13+(13-1)*10 =133
a14 =a1+(n-1)*d =13+(14-1)*10 =143
a15 =a1+(n-1)*d =13+(15-1)*10 =153
a16 =a1+(n-1)*d =13+(16-1)*10 =163
a17 =a1+(n-1)*d =13+(17-1)*10 =173
a18 =a1+(n-1)*d =13+(18-1)*10 =183
a19 =a1+(n-1)*d =13+(19-1)*10 =193
a20 =a1+(n-1)*d =13+(20-1)*10 =203
a21 =a1+(n-1)*d =13+(21-1)*10 =213
a22 =a1+(n-1)*d =13+(22-1)*10 =223
a23 =a1+(n-1)*d =13+(23-1)*10 =233
a24 =a1+(n-1)*d =13+(24-1)*10 =243
a25 =a1+(n-1)*d =13+(25-1)*10 =253
a26 =a1+(n-1)*d =13+(26-1)*10 =263
a27 =a1+(n-1)*d =13+(27-1)*10 =273
a28 =a1+(n-1)*d =13+(28-1)*10 =283
a29 =a1+(n-1)*d =13+(29-1)*10 =293
a30 =a1+(n-1)*d =13+(30-1)*10 =303
a31 =a1+(n-1)*d =13+(31-1)*10 =313
a32 =a1+(n-1)*d =13+(32-1)*10 =323
a33 =a1+(n-1)*d =13+(33-1)*10 =333