Question

How many two-digit natural numbers are divisible by $7$?

Answer 1

Step 1: Identify the given sequence as an arithmetic progression

The first two digit natural number divisible by $7$ is $14$

The last two digit natural number divisible by $7$is $98$

The sequence of numbers $14,21,28,\dots .,98$ which is divisible by $7$ is an arithmetic progression with$1^{st}$ term $14$ and common difference of $7$

Step 2 : Finding the number of terms in an arithmetic progression

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

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

Here,$a_n=98,a=14,d=7$

$$\begin{gathered} \Rightarrow 98=14+(n-1)7 \\ \Rightarrow 84=7n-7 \\ \Rightarrow 7n=91 \\ \Rightarrow n=13 \end{gathered}$$

Hence, the number of two digit natural numbers divisible by $7$ is $13$