Question 1185343
this looks like an arithmetic sequence type problem.


in an arithmetic sequence An = A1 + (n-1) * d


if d = 4, then An = A1 + (n-1) * 4


A2 would be equal to A1 + 1 * 4 which is equal to A1 + 4.


if A1 is equal to 10, then A2 would be equal to 14.


A3 would be equal to 10 + 2 * 4 = 18 which is also equal to A2 + 4


A4 would be equal to 10 + 3 * 4 = 22 which is also equal to A3 + 4


what this is saying is that An+1 is always equal to An + 4.


A20 would be equal to 10 + 19 * 4 = 10 + 76 = 86


the sum of an arithmetic sequence is equal to n/2 * (A1 + An).


since An = A1 + (n-1) * d, this formula can also be expressed as:


sum = n/2 * (A1 + A1 + (n-1) * d).


this becomes sum = n/2 * (2 * A1 + (n-1) * d).


either way, you'll get the sum.


if sum = n/2 * (A1 + An), you first have to find An.


when n = 20 and A1 = 10 and d = 4, A20 = 10 + 19 * 4 =  10 + 76 = 86.


sum is then equal to 20/2 * 10 + 86) = 10 * 96 = 960.


if sum = n/2 * (2 * A1 + (n-1) * d), then:


sum = 20/2 * (2 * 10 + 19 * 4) which becomes equal to 10 * (20 + 76) which is equal to 10 * 96 which is equal to 960.


the two formulas get you the same answer.


if you simply did the sum manually, you would get:


A1 = 10
A2 = 14
A3 = 18
A4 = 22
A5 = 26
A6 = 30
A7 = 34
A8 = 38
A9 = 42
A10 = 46
A11 = 50
A12 = 54
A13 = 58
A14 = 62
A15 = 66
A16 = 70
A17 = 74
A18 = 78
A19 = 82
A20 = 86


add those up and you get 960.