Question

sum or intégrés from 11to 1000 divisible by3 explanation

Answer 1

1st integer divisible by 3 between 11 and 1000= 12
last integer divisible between 11 and 1000= 999
common difference betwwn 2 integers= 3
hence, it forms an AP
last term=1st term+ [no. of terms-1}common difference
999=12+(n-1)3
n-1= 329
n = 330
sum of terms of an AP= n/2(1st term+ last term)
                                    =330/2(12+999)
                                    = 165*1011
                                    = 166815
ans. 166815