find the sum of all three digit natural numbers divisible by 7????
The list of 3 digit numbers divisible by 7 are :
105, 112, ........994.
The above list forms an AP with first term, a = 105 and common difference, d = 7
Let an = 994
so, a + (n - 1)d = 994
⇒ 105 + (n - 1)7 = 994
⇒ n = 128
now, sum of all the 3 digit numbers divisible by 7 is given by,
S128 = (128/2) [2×105 +(128 - 1)7]
= 64 [210 + 889] = 64 × 1099 = 70336