find the sum of all three digit natural numbers which are divisible by `7`

Explanation

The smallest and the largest numbers of three digits, which are divisble by 7 are 105 and 994, respectively. So, the sequence of three digit numbers which are divisble by 7 is 105,112,119,…,994. Clearly, it is an A.P. with first term a=105 and common difference d=7. Let there be n terms in this sequence.
Then,
`a_(n)=994`
or a+(n-1)d=994
or 105+(n-1)`xx7=994`
or n=128
Now, Required sum =`n/2[2a+(n-1)d]`
`=128/2[2xx105+(128-1)xx7]`
=70336