Question

find the sum of the sequence 16,23,30...93?​

Answer 1

Answer:

a2-a1=23-16=7

a3-a2=30-23=7

a4-a3=37-30=7

The difference between every two adjacent members of the series is constant and equal to 7

a1=16 (this is the 1st member)

an=37 (this is the last/nth member)

d=7 (this is the difference between consecutivemembers)

n=4 (this is the number of members)

16+23+30

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 16 + 37 = 53), and dividing by 2:

n(a1+an)2

4(16+37) 2

The sum of the 4 members of this series is 106

This series corresponds to the following straight line y=7x+16

a1 =a1+(n-1)*d =16+(1-1)*7 =16

a2 =a1+(n-1)*d =16+(2-1)*7 =23

a3 =a1+(n-1)*d =16+(3-1)*7 =30

a4 =a1+(n-1)*d =16+(4-1)*7 =37

a5 =a1+(n-1)*d =16+(5-1)*7 =44

a6 =a1+(n-1)*d =16+(6-1)*7 =51

a7 =a1+(n-1)*d =16+(7-1)*7 =58

a8 =a1+(n-1)*d =16+(8-1)*7 =65

a9 =a1+(n-1)*d =16+(9-1)*7 =72

a10 =a1+(n-1)*d =16+(10-1)*7 =79

a11 =a1+(n-1)*d =16+(11-1)*7 =86

a12 =a1+(n-1)*d =16+(12-1)*7 =93