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.
