Question

find the sum of all odd number divisible by 3 between 1 and 1000

Answer 1

hey friend,

Let the numbers are 3,6,9,12,15,..................999 

If the initial term of an arithmetic progression is a1 (= 3 here) and the common difference of successive members is d (= 3 here), then the nth term of the sequence is given by: 
a(n) = a(1) + (n - 1)*d 
999 = 3 + (n - 1)*3 
996/3 = n - 1 
332= n - 1 or n = 333 terms 
Now sum of the progression with first term is a(1) = 3 and number of term (n) = 333 and last term is a(n) = 999 
S(n) = n/2*[a(1) + a(n)] 
S(n) = 333/2*[3 + 999] 
S(n) = (333/2)*(1002)
S(n) = 166833. ................. Answer 

Hope this helped