Question

Find, how many two digit natural numbers are divisible by 7.

Answer 1

The first two-digit number that is divisible by 7 is 14

So, the sequence starts from 14

And the highest two-digit number that is divisible by 7 is 98

So, the sequence becomes:
14, 21, ..., 98.
We, need to find the numbers in the given sequence
Using a_n = a + (n - 1)d

a is the first term, d is the common difference, n is the number of terms and  is the n^th term and a_n is the nth term 
a = 14, d = 7, an= 98 substituting the values we get 

98 = 14 + (n - 1)7

98 = 14 + 7n - 7

91 = 7n

n = `91/7`

n = 13

Therefore, there are total 13 two digit numbers that are divisible by 7.