How many three digit numbers are divisible by 7?

We have AP starting from 105 because it is the first three digit number divisible by 7.

AP will end at 994 because it is the last three digit number divisible by 7.

Therefore, we have AP of the form 105, 112, 119…, 994

Let 994 is the \(n^{th}\) term of AP.

We need to find n here.

First term = a = 105,
Common difference = d
= 112 – 105 = 7

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 994 = 105 + (n - 1) (7)
\(\Rightarrow \) 994 = 105 + 7n - 7
\(\Rightarrow \) 896 = 7n
\(\Rightarrow \) n = 128

It means 994 is the \(128^{th}\) term of AP.

Therefore, there are 128 terms in AP.