Question

the sum of all natural numbers between 100and 1000 which arw multiple of 5 is :​

Answer 1

Answer:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995. This sequence forms an A.P. Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

a=100

an=1000

d=5

n=(an-a/d)+1

n=181

sn=n/2(2a+(n-1)d)

=181/2(200+180×5)

=181/2(1100)

=181×550

=98450