**Answer:**

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

This sequence forms an A.P.

Here, first term, a = 105

Common difference, d = 5

Here,

a+(n−1)d = 995

=> 105+(n−1)5 = 995

=> (n−1)5 = 995−105 = 890

=> n−1 = 178

=> n = 179

Sn = n/2[2a+(n−1)d]

∴ Sn = 179/2[2×(105)+(179−1)×(5)]

= 179/2[2(105)+(178)(5)]

= 179[105+(89)5]

= (179)[105+445]

= 179×550

= 98450

**Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.**