# How many three-digit numbers are divisible by 7?

**Hint:**An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……

**Complete step-by-step answer:**

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

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 , to find ${n^{th}}$ term of arithmetic progression,

\[\begin{array}{*{20}{l}} { \Rightarrow 994{\text{ }} = {\text{ }}105{\text{ }} + {\text{ }}\left( {n{\text{ }} - {\text{ }}1} \right){\text{ }}\left( 7 \right)} \\ { \Rightarrow 994{\text{ }} = {\text{ }}105{\text{ }} + {\text{ }}7n - 7} \\ \Rightarrow 896{\text{ }} = {\text{ }}7n \\ \Rightarrow n{\text{ }} = {\text{ }}128 \\ \end{array}\]

It means 994 is the ${128^{th}}$ term of AP.

Therefore, there are 128 terms in AP.

**Hence, 128 three digit numbers are divisible by 7.**

**Note:** To solve most of the problems related to AP, the terms can be conveniently taken as

3 terms: (a−d),a,(a+d)

4 terms: (a−3d),(a−d),(a+d),(a+3d)

5 terms: (a−2d),(a−d),a,(a+d),(a+2d)

${t_n}={S_n}-{S_{n-1}}$

If each term of an AP is increased, decreased, multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.

In an AP, the sum of terms equidistant from beginning and end will be constant.