Q. 24 B4.8( 4 Votes )

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

Answer :

The numbers lying between 100 and 1000 that are multiples of 5 are


105, 110, 115, 120,…, 995


a2 – a1 = 110 – 105 = 5


a3 – a2 = 115 – 110 = 5


a3 – a2 = a2 – a1 = 5


Therefore, the series is in AP


Here, a = 105, d = 5 and an = 995


We know that,


an = a + (n – 1)d


995 = 105 + (n – 1)5


995 – 105 = (n – 1)5


890 = (n – 1)5


178 = (n – 1)


n = 179


Now, we have to find the sum of this AP




S179 = 179[105 + 89 × 5]


S179 = 179 [550]


S179 = 98450


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


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