Find the sum of all the three digit natural numbers which are multiples of 7.

The three digit natural numbers which are multiples of 7 are

105, 112, 119,…, 994

a₂ – a₁ = 112 – 105 = 7

a₃ – a₂ = 112 – 105 = 7

∵ a₃ – a₂ = a₂ – a₁ = 7

Therefore, the series is in AP

Here, a = 105, d = 7 and a_n = 994

We know that,

a_n = a + (n – 1)d

⇒ 994 = 105 + (n – 1)7

⇒ 994 – 105 = (n – 1)7

⇒ 889 = (n – 1)7

⇒ 127 = (n – 1)

⇒ n = 128

Now, we have to find the sum of this AP

⇒ S₁₂₈ = 64[210 + 127 × 7]

⇒ S₁₂₈ = 64[1099]

⇒ S₁₂₈ = 70336

Hence, the sum of all three digit numbers which are multiples of 7 are 70336.