Find the sum of all 3 digit natural multiples of 6.

Now the first 3 digit multiple of 6 is 102.

That means, a = 102

And all the nos. Are multiples of 6 which means that they have a difference of 6 in between them so, common difference d = 6

Now, the last 3 digit multiple of 6 is 996.

Now, we know that, T_n = a + (n – 1)d

So, 996 = a + (n – 1)d

⇒ 996 = 102 + (n – 1)6

⇒ 6(n – 1) = 996 – 102

⇒ 6(n – 1) = 894

⇒ n – 1 =

⇒ n – 1 = 149

⇒ n = 150

So, we can say that there are 150, 3 digit multiples of 6

Now sum of these nos. S₁₅₀ = × (2a + (n – 1)d)

⇒ S₁₅₀ = × (2(102) + (150 – 1)6)

= 75 × (204 + 149 × 6)

= 75 × (1098)

⇒ S₁₅₀ = 82350

∴ Sum of all 3 digit natural multiples of 6 is 82350.