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Q.
The sum of all three digit natural numbers, which are divisible by $7$ is
Sequences and Series
Solution:
The smallest and the largest numbers of three digits, which are divisible by $7$ are $105$ and $994$ respectively. So, the sequence of three digit numbers which are divisible by $7$ is $105,112,119, \ldots, 994 .$ 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 $
$ \Rightarrow a+(n-1) d=994 $
$\Rightarrow 105+(n-1) \times 7=994$
$\Rightarrow n=128 $
Now, required sum $=\frac{n}{2}\left[a_{1}+a_{n}\right]$
$=\frac{128}{2}[105+994]=70336$