Question

Show that the sum of all odd integer between 1 and 1000 which are divisible by 3 is 83667

Answer 1

Answer:

Step-by-step explanation:

The numbers are 3, 9, 15, 21,..................999 

If the initial term of an arithmetic progression is a1 (= 3 here) and the common difference of successive members is d (= 6 here), then the nth term of the sequence is given by: 

a(n) = a(1) + (n - 1)*d 

999 = 3 + (n - 1)*6 

996/6 = n - 1 

166 = n - 1 or n = 167 terms 

Now sum of the progression with first term is a(1) = 3 and number of term (n) = 167 and last term is a(n) = 999 

S(n) = n/2*[a(1) + a(n)] 

S(n) = 167/2*[3 + 999] 

S(n) = (167/2)*(1002) or 167*501 

S(n) = 83,667 ................. Answer