Question

The sum of all 3-digit natural numbers which are divisible by 13 is

Answer 1

The three digit natural numbers which are divisible by 13 are 104,117,130,.............,988

Here, a = 104 , d = 13 , tn = 988

we know that, tn = a + ( n - 1 ) d
Therefore, 988 = 104 + ( n - 1 ) 13
988 = 104 + 13n - 13
13n = 988-104+13
13n = 988 - 91
13n = 897
n = 897÷13
n = 69.....

now the sum of all 3-digits natural numbers divisible by 13
a = 104 , d = 13 , n = 69

we know that, Sn = n÷2+[ 2a + ( n - 1 ) d ]
Therefore, S-69 = 69÷2+ [ 2×104 + ( 69-1 ) × 13 ]
S-69 = 69÷2+ [ 208 + 68 ×13 ]
S-69 = 69÷2+ [208+884]
S-69 = 69÷2+1092
S-69 = 34.5 + 1092
S-69 = 1126.5.......

Therefore, the sum of all 3 digit natural numbers divisible by 13 is 1126.5