3+8+13+18+...+58 gauss’s approach to find the following sum
Answers
Answer:
Step-by-step explanation:
3,8,13,18
Your input 3,8,13,18 appears to be an arithmetic sequence
Find the difference between the members
a2-a1=8-3=5
a3-a2=13-8=5
a4-a3=18-13=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
=3+(n-1)5
a1=3 (this is the 1st member)
an=18 (this is the last/nth member)
d=5 (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:
3+8+13+18
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 3 + 18 = 21), and dividing by 2:
n(a1+an)
2
4(3+18)
2
The sum of the 4 members of this series is 42
This series corresponds to the following straight line y=5x+3
Finding the n
th
element
a1 =a1+(n-1)*d =3+(1-1)*5 =3
a2 =a1+(n-1)*d =3+(2-1)*5 =8
a3 =a1+(n-1)*d =3+(3-1)*5 =13
a4 =a1+(n-1)*d =3+(4-1)*5 =18
a5 =a1+(n-1)*d =3+(5-1)*5 =23
a6 =a1+(n-1)*d =3+(6-1)*5 =28
a7 =a1+(n-1)*d =3+(7-1)*5 =33
a8 =a1+(n-1)*d =3+(8-1)*5 =38
a9 =a1+(n-1)*d =3+(9-1)*5 =43
a10 =a1+(n-1)*d =3+(10-1)*5 =48
a11 =a1+(n-1)*d =3+(11-1)*5 =53
a12 =a1+(n-1)*d =3+(12-1)*5 =58
a13 =a1+(n-1)*d =3+(13-1)*5 =63
a14 =a1+(n-1)*d =3+(14-1)*5 =68
a15 =a1+(n-1)*d =3+(15-1)*5 =73
a16 =a1+(n-1)*d =3+(16-1)*5 =78
a17 =a1+(n-1)*d =3+(17-1)*5 =83
a18 =a1+(n-1)*d =3+(18-1)*5 =88
a19 =a1+(n-1)*d =3+(19-1)*5 =93
a20 =a1+(n-1)*d =3+(20-1)*5 =98
a21 =a1+(n-1)*d =3+(21-1)*5 =103
a22 =a1+(n-1)*d =3+(22-1)*5 =108
a23 =a1+(n-1)*d =3+(23-1)*5 =113
a24 =a1+(n-1)*d =3+(24-1)*5 =118
a25 =a1+(n-1)*d =3+(25-1)*5 =123
a26 =a1+(n-1)*d =3+(26-1)*5 =128
a27 =a1+(n-1)*d =3+(27-1)*5 =133
a28 =a1+(n-1)*d =3+(28-1)*5 =138
a29 =a1+(n-1)*d =3+(29-1)*5 =143
i think it helps you!