Question

How many three digit numbers are divisible by $11$? Also find the middle term.

Answer 1

Step 1: Identify the given sequence in an arithmetic progression

The first three digit number divisible by $11$ is $110$

The last three digit number divisible by $11$is $990$

The sequence of numbers $110,121,132,......,990$ which is divisible by 11 is an arithmetic progression with $1^{st}$ term $110$ and common difference of $11$

Step 2: Find the number of terms in the arithmetic progression

The $n^{th}$ term of an arithmetic progression is given by

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

Here, $a_n=990,a=110,d=11$

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

Step 3: Find the middle term

Since the sequence has $81$ terms, the ${41}^{st}$ term is the middle term

$$\begin{gathered} \Rightarrow a_{41}=a+(41-1)d \\ \Rightarrow a_{41}=110+(41-1)11 \\ \Rightarrow a_{41}=110+40\times 11 \\ \Rightarrow a_{41}=550 \end{gathered}$$

Hence, there are $81$ three digit numbers divisible by $11$and the middle term of the sequence is $550$.