can you solve it 1) 6 11 16 21 26 ? ?
Answers
Answer:
main difference 5
next step:- 31...36.....and so on
Answer:
hope u help this
Step-by-step explanation:
Your input 1,6,11,16,21,26,31,36 appears to be an arithmetic sequence
Find the difference between the members
a2-a1=6-1=5
a3-a2=11-6=5
a4-a3=16-11=5
a5-a4=21-16=5
a6-a5=26-21=5
a7-a6=31-26=5
a8-a7=36-31=5
The difference between every two adjacent members of the series is constant and equal to 5
General Form: a
n
=a
1
+(n-1)d
a
n
=1+(n-1)5
a1=1 (this is the 1st member)
an=36 (this is the last/nth member)
d=5 (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:
1+6+11+16+21+26+31+36
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 1 + 36 = 37), and dividing by 2:
n(a1+an)
2
8(1+36)
2
The sum of the 8 members of this series is 148
This series corresponds to the following straight-linen y=5x+1
Finding the n
th
element
a1 =a1+(n-1)*d =1+(1-1)*5 =1
a2 =a1+(n-1)*d =1+(2-1)*5 =6
a3 =a1+(n-1)*d =1+(3-1)*5 =11
a4 =a1+(n-1)*d =1+(4-1)*5 =16
a5 =a1+(n-1)*d =1+(5-1)*5 =21
a6 =a1+(n-1)*d =1+(6-1)*5 =26
a7 =a1+(n-1)*d =1+(7-1)*5 =31
a8 =a1+(n-1)*d =1+(8-1)*5 =36
a9 =a1+(n-1)*d =1+(9-1)*5 =41
a10 =a1+(n-1)*d =1+(10-1)*5 =46
a11 =a1+(n-1)*d =1+(11-1)*5 =51
a12 =a1+(n-1)*d =1+(12-1)*5 =56
a13 =a1+(n-1)*d =1+(13-1)*5 =61
a14 =a1+(n-1)*d =1+(14-1)*5 =66
a15 =a1+(n-1)*d =1+(15-1)*5 =71
a16 =a1+(n-1)*d =1+(16-1)*5 =76
a17 =a1+(n-1)*d =1+(17-1)*5 =81
a18 =a1+(n-1)*d =1+(18-1)*5 =86
a19 =a1+(n-1)*d =1+(19-1)*5 =91
a20 =a1+(n-1)*d =1+(20-1)*5 =96
a21 =a1+(n-1)*d =1+(21-1)*5 =101
a22 =a1+(n-1)*d =1+(22-1)*5 =106
a23 =a1+(n-1)*d =1+(23-1)*5 =111
a24 =a1+(n-1)*d =1+(24-1)*5 =116
a25 =a1+(n-1)*d =1+(25-1)*5 =121
a26 =a1+(n-1)*d =1+(26-1)*5 =126
a27 =a1+(n-1)*d =1+(27-1)*5 =131
a28 =a1+(n-1)*d =1+(28-1)*5 =136
a29 =a1+(n-1)*d =1+(29-1)*5 =141
a30 =a1+(n-1)*d =1+(30-1)*5 =146
a31 =a1+(n-1)*d =1+(31-1)*5 =151
a32 =a1+(n-1)*d =1+(32-1)*5 =156
a33 =a1+(n-1)*d =1+(33-1)*5 =161