Question

find the sum of integers from 11 to 1000 which are divisible by 3

Answer 1

a=12,a+d=15,
the ap is in the form 12,15,18,21....999
n=an-a/d+1
n=999-12/3+1
n=987/3+1
n=369+1
n=370
sn=n/2× (an-a)/d
sn=370/2×999-12/3
sn=185×329

Answer 2

the sum of the multiples of 3 between 11 to 1000 is Sn=12+15+18+...+999
initial term a1=12
common difference d=3

then the nth term of the
sequence is given by:
a(n) = a(1) + (n - 1)*d
999 = 12 + (n - 1)*3
987/3=n-1
329= n - 1 or n = 330 terms
Now sum of the progression with first term is a(1) = 12
and number of term (n) = 330and last term is a(n) =999
S(n) = n/2*[a(1) + a(n)]
S(n) = 330/2*[12+ 999]
S(n) =165 *1011=1661815