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 a_{n }= 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,

S_{128 }= (128/2) [2×105 +(128 - 1)7]

= 64 [210 + 889] = 64 × 1099 = 70336

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