Question

Find the sum of all 3- digit natural numbers which are divisible by $13$.

Answer 1

first natural no be $\to 104$

second natural no be $\to 117$

last natural no. be $\to 988$

$a_1=104,d=a_2-a_1=117-104$

$a_2=117$

$d=13,a_n=988$

$\Rightarrow a_n=a+(n-1)d$

$\Rightarrow 988=104+(n-1)13$

$\Rightarrow \frac{988-104}{13}=n-1$

$\Rightarrow \frac{884}{13}=n-1$

$n=69$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right],n=69$

$S_{69}=\frac{69}{2}[2\times 104+(69-1)13$

$S_{69}=\frac{69}{2}[208+68\times 13]$

$S_{69}=\frac{69}{2}[208+884]$

$=\frac{69}{2}\times 1092$

So, sum of all natural numbers $=37674$

$S_{69}=37674$