AP ICET 2019 Shift 1 Paper

Question : 129 of 200

Marks: +1, -0

The sum of all the natural numbers between 100 and 1000 which are multiples of 5 is

Solution:

The multiples of 5 between 100 and 1000 are 105, 110, ----- 995 which are in A.P.
Where

a= 105, d = 5, t_n = 995

a + (n – 1) d = 995⇒n = 179

∴S_n={179/2}(105+995)=98450

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