Question

2.Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
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Answer 1

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=n2[2a+(n−1)d]

 

∴Sn=1792[2×(105)+(179−1)×(5)]

 

=1792[2(105)+(178)(5)]

 

=179[105+(89)5]

 

=(179)[105+445]

 

=179×550

 

=98450

 

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