The numbers lying between 100 and 1000 that are multiples of 5 are
105, 110, 115, 120,…, 995
a2 – a1 = 110 – 105 = 5
a3 – a2 = 115 – 110 = 5
∵ a3 – a2 = a2 – a1 = 5
Therefore, the series is in AP
Here, a = 105, d = 5 and an = 995
We know that,
an = a + (n – 1)d
⇒ 995 = 105 + (n – 1)5
⇒ 995 – 105 = (n – 1)5
⇒ 890 = (n – 1)5
⇒ 178 = (n – 1)
⇒ n = 179
Now, we have to find the sum of this AP
⇒ S179 = 179[105 + 89 × 5]
⇒ S179 = 179 [550]
⇒ S179 = 98450
Hence, the sum of all numbers lying between 100 and 1000 that are multiples of 5 is 98450.