Question

Show that the sum of all odd integers between 1and 100 which are divisible by 3 is 83667

Answer 1

HEY DEAR ... ✌️

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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 

HOPE , IT HELPS ... ✌️