Find the sum of all two digit natural numbers which are divisible by 4.

The sum of series of the two digit natural numbers that are divisible by 4 is $12 + 16 + 20 + …. + 96.$

a = 12, d = 16 - 12 = 4 and l = 96.

Let 96 be nth term then,

⇒ 96 = 12 + 4(n - 1)
⇒ 96 - 12 = 4n - 4
⇒ 84 = 4n - 4
⇒ 4n = 84 + 4
⇒ 4n = 88
⇒ n = 22.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₂₂ = $\dfrac{22}{2}[2 \times 12 + 4(22 - 1)]$
⇒ S₂₂ = 11[24 + 84]
⇒ S₂₂ = 11 × 108
⇒ S₂₂ = 1188.

Hence, the sum of all two digits natural numbers which are divisible by 4 is 1188.