Find the sum of all 3 digit natural numbers which are divisible by 13
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104, 117, 130........... 988
a = 104
d = 13
L = 988
=> a + (n-1) d = 988
=> 104 + (n-1) (13) = 988
=> (n-1) (13) = 988 - 104
=> (n-1) (13) = 884
=> n-1 = 68
=> n = 69
S69 = 69/2 [ 2 × 104 + (69 - 1)13]
= 69 /2 [ 208 + 68 × 13]
= 69 /2 [ 208 + 884]
= 69 × 1092 / 2
= 69 × 546
= 37674
a = 104
d = 13
L = 988
=> a + (n-1) d = 988
=> 104 + (n-1) (13) = 988
=> (n-1) (13) = 988 - 104
=> (n-1) (13) = 884
=> n-1 = 68
=> n = 69
S69 = 69/2 [ 2 × 104 + (69 - 1)13]
= 69 /2 [ 208 + 68 × 13]
= 69 /2 [ 208 + 884]
= 69 × 1092 / 2
= 69 × 546
= 37674
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