Question

Find the sum of all natural numbers between $1$ and $201$ which are divisible by $5$.

Answer 1

Since the sequence is the sum of all natural numbers between $41$ and $201$, therefore, the sequence is of the type $5+10+\mathrm{15.........}+200$

We know the $n$th term of an A.P with first term $a$ and the common difference $d$ is:

$t_n=a+(n-1)d$

Here, the first term is $a=5$, common difference is $d=10-5=5$ and $t_n=200$, therefore,

$t_n=a+(n-1)d\Rightarrow 200=5+(n-1)5\Rightarrow 200=5+5n-5\Rightarrow 5n=200\Rightarrow n=\frac{200}{5}=40$

We also know the sum of $n$ terms of an A.P with first term $a$ and the common difference $d$ is:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Here, the first term is $a=5$, number of terms $n=40$ and common difference is $d=5$, therefore,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{40}{2}\left[\right(2\times 5)+(40-1\left)\right(5\left)\right]=20(10+195)=20\times 205=4100$

Hence, the sum of all natural numbers between $41$ and $201$ is $4100$.