# Find the sum of all natural numbers lying between 100 and 500, which are divisible by 8.

The numbers lying between 100 and 500 which are divisible by 8 are

104, 112, 120, 128, 136,…, 496

a2 – a1 = 112 – 104 = 8

a3 – a2 = 120 – 112 = 8

a3 – a2 = a2 – a1 = 8

Therefore, the series is in AP

Here, a = 120, d = 8 and an = 496

We know that,

an = a + (n – 1)d

496 = 104 + (n – 1)8

496 – 104 = (n – 1)8

392 = (n – 1)8

49 = (n – 1)

n = 50

Now, we have to find the sum of this AP

S50 = 25[208 + 49 × 8]

S50 = 25[600]

S50 = 15000

Hence, the sum of all numbers lying between 100 and 500 and divisible by 8 is 15000.

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