Question

Find the sum of all three-digit natural numbers, which are multiples of $11$.

Answer 1

Step 1: Find the first term, common difference and $\mathit{n}$:

Formula:

The $n^{\mathrm{th}}$ term of an AP is $a_n=a+\left(n-1\right)d$.

All three-digit natural numbers that are multiples of $11$ are $110,121,132,\dots ,990$.

Here, the first term, $a=110$, the common difference, $d=121-110$, that is, $d=11$ and the $n^{\mathrm{th}}$ term is, $a_n=990$. Substitute these values in the $n^{\mathrm{th}}$ term of an AP formula.

$$\begin{gathered} \Rightarrow 990=110+(n-1)11 \\ \Rightarrow 880=11n-11 \\ \Rightarrow 11n=891 \\ \Rightarrow n=81 \end{gathered}$$

Step 2: Evaluate the sum of $\mathbf{81}$ terms:

Formula:

The sum of $n$ terms of a series in an AP is $S_n=\frac{n}{2}\left[a+a_n\right]$.

Substitute $a=110$, $a_n=990$ and $n=81$ into the $S_n$ formula.

$\begin{array}{rcl}& \Rightarrow & S_{81}=\frac{81}{2}\left(110+990\right)\\ & =& \frac{81}{2}\times 1100\\ & =& 44550\end{array}$

Hence, the required sum is $S_{81}=44550$.