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.