Question

Find the sum of all natural numbers between $300$ and $500$ which are divisible by $11$

Answer 1

The natural numbers between $300$ and $500$ which are divisible by $11$ are as follows:

$308,319,330,.....,495$

In the above sequence, the first term is $a_1=308$, second term is $a_2=319$ and the $n$th term is $T_n=495$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=319-308=11$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, with $a=308,d=11$ and $T_n=495$, we have

$T_n=a+(n-1)d\Rightarrow 495=308+(n-1)11\Rightarrow 495=308+11n-11\Rightarrow 495=297+11n\Rightarrow 11n=495-297\Rightarrow 11n=198\Rightarrow n=\frac{198}{11}\Rightarrow n=18$

We also know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now to find the sum of series, substitute $n=18,a=308$ and $d=11$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{18}=\frac{18}{2}\left[\right(2\times 308)+(18-1\left)11\right]=9[616+(17\times 11\left)\right]=9(616+187)=9\times 803=7227$

Hence, the sum is $7227$.