Find the sum of all three-digit natural numbers which are divisible by 13.
Let the number of terms be n
Then Tn=988
a+(n−1)d=988
104+(n−1)13=988
(n−1)13=988−104
n−1=88413
n−1=68
n=68+1
∴n=69
Therefore, the required sum =n2(a+l)
=692(104+988)
=69×10922
=69×546
=37674
Hence, the required sum is 37674.