Question -
Answer -
The odd numbersbetween 1 and 1000 divisible by 3 are 3, 9, 15,…,999
Let the number ofterms be ‘n’, so the nth term is 999
a = 3, d = 9-3 = 6, an =999
an = a+ (n-1)d
999 = 3 + (n-1)6
999 = 3 + 6n – 6
6n = 999 + 6 – 3
6n = 1002
n = 1002/6
= 167
By using the formula,
Sum of n terms, S =n/2 [a + l]
= 167/2 [3 + 999]
= 167/2 [1002]
= 167 [501]
= 83667
∴ The sum of all oddintegers between 1 and 1000 which are divisible by 3 is 83667.
Hence proved.