1 question

1) The first multiple of 5 is 5, the last multiple is 200.a) Required no. = ⅕(200 – 5) + 1 = 40b) Required sum = ½(200 + 5)(40) = 4100c) Required sum = ½(200 + 1)(200) – 4100 = 16000 I HOPE THIS CAN HELP YOU! ~ ^_^

2012-07-28 19:16:43 補充：
Formula:
i) no. of terms = (last term - first term)/(common difference) + 1
ii) sum of terms = (first term + last term)(no. of terms)/2

(a) For the positive integers 1 to 200 inclusive, the 1st multiple of 5 is 5X1=5, the last multiple of 5 is 5X40=200, so the number of all the multiples of 5 is 40.

(b) the sum of all the multiples of 5 is:
5X1 + 5X2 + 5X3 + ...... + 5X40
=5X(1+2+3+......+40)
=5X(1+40)X40/2
=4100

(c) the sum of all the integers wich are not the multiples of 5 is:
(1+2+3+4+......+200)-4100
=(1+200)x200/2-4100
=16000