Answer :

Three - digit natural numbers which are divisible by 13 are 104, 117, 130, …, 988.

Sum of these numbers forms an arithmetic series 104 + 117 + 130 + … + 988.


Here, first term = a = 104


Common difference = d = 13


We first find the number of terms in the series.


Here, last term = l = 988


988 = a + (n - 1)d


988 = 104 + (n - 1)13


988 - 104 = 13n - 13


884 = 13n - 13


884 + 13 = 13n


13n = 897


n = 69


Now, Sum of n terms of this arithmetic series is given by:


Sn = [2a + (n - 1)d]


Therefore sum of 69 terms of this arithmetic series is given by:


S69 = [2(104) + (69 - 1)(13)]


= (69/2) × [208 + 884]


= (69/2) × 1092


= 69 × 546


= 3767


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