Question:
Find the sum of all three-digit natural numbers which are divisible by 13.
Solution:
All three-digit numbers which are divisible by 13 are 104, 117, 130, 143,..., 988.
This is an AP in which a = 104, d = (117 - 104) = 13 and l = 988
Let the number of terms be n.
Then Tn = 988
⇒ a + (n - 1)d = 988
⇒ 104 + (n -1) ⨯ 13 = 988
⇒ 13n = 897
⇒ n = 69
$\therefore$ Required sum $=\frac{n}{2}(a+l)$
$=\frac{69}{2}[104+988]=69 \times 546=37674$
Hence, the required sum is 37674.
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