Question

Find the algebra of the sum
ń term
of the sequence 6, 10, 14? ​

Answer 1

Answer:

a1 =a1+(n-1)*d =6+(1-1)*4 =6

a2 =a1+(n-1)*d =6+(2-1)*4 =10

a3 =a1+(n-1)*d =6+(3-1)*4 =14

a4 =a1+(n-1)*d =6+(4-1)*4 =18

a5 =a1+(n-1)*d =6+(5-1)*4 =22

a6 =a1+(n-1)*d =6+(6-1)*4 =26

a7 =a1+(n-1)*d =6+(7-1)*4 =30

a8 =a1+(n-1)*d =6+(8-1)*4 =34

a9 =a1+(n-1)*d =6+(9-1)*4 =38

a10 =a1+(n-1)*d =6+(10-1)*4 =42

a11 =a1+(n-1)*d =6+(11-1)*4 =46

a12 =a1+(n-1)*d =6+(12-1)*4 =50

a13 =a1+(n-1)*d =6+(13-1)*4 =54

a14 =a1+(n-1)*d =6+(14-1)*4 =58

a15 =a1+(n-1)*d =6+(15-1)*4 =62

a16 =a1+(n-1)*d =6+(16-1)*4 =66

a17 =a1+(n-1)*d =6+(17-1)*4 =70

a18 =a1+(n-1)*d =6+(18-1)*4 =74

a19 =a1+(n-1)*d =6+(19-1)*4 =78

a20 =a1+(n-1)*d =6+(20-1)*4 =82

a21 =a1+(n-1)*d =6+(21-1)*4 =86

a22 =a1+(n-1)*d =6+(22-1)*4 =90

a23 =a1+(n-1)*d =6+(23-1)*4 =94

a24 =a1+(n-1)*d =6+(24-1)*4 =98

a25 =a1+(n-1)*d =6+(25-1)*4 =102

a26 =a1+(n-1)*d =6+(26-1)*4 =106

a27 =a1+(n-1)*d =6+(27-1)*4 =110

a28 =a1+(n-1)*d =6+(28-1)*4 =114

a29 =a1+(n-1)*d =6+(29-1)*4 =118

Step-by-step explanation:

a2-a1=10-6=4

a3-a2=14-10=4

a4-a3=18-14=4

The difference between every two adjacent members of the series is constant and equal to 4