# Ex.5.2 Q13 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

How many three-digit numbers are divisible by \(7\)?

## Text Solution

**What is Known:?**

Three-digit numbers

**What is Unknown?**

Number of all three-digit numbers which are divisible by \(7\)

**Reasoning:**

\({a_n} = a + \left( {n - 1} \right)d\) is the general term of AP. Where \({a_n}\) is the \(n \rm{th}\) term, \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

First three-digit number that is divisible by \(7 = 105\)

Next number \(= 105 + 7 = 112\)

Therefore, the series becomes \(105,\, 112, \,119,\,\dots\)

All are three-digit numbers which are divisible by \(7\) and thus, all these are terms of an A.P. having first term as \(105\) and common difference as \(7.\)

When we divide \(999\) by \(7,\) the remainder will be \(5.\)

Clearly, \(999 − 5 = 994\) is the maximum possible three-digit number that is divisible by \(7.\)

Hence the final series is as follows:

\[105, \,112,\, 119,\, \dots, 994\]

Let \(994\) be the \(n\rm{th} \) term of this A.P.

\[\begin{align}a &= 105\\d &= 7\\an &= 994\\n &= ?\end{align}\]

We know that the nth term of an A.P. Series,

\[\begin{align}{a_n} &= a + (n - 1)d\\994 &= 105 + (n - 1)7\\

889 &= (n - 1)7\\n - 1 &= \frac{{889}}{7}\\n - 1& = 127\\{n}& = 127 + 1\\n &= 128\end{align}\]

Therefore, \(128\) three-digit numbers are divisible by \(7.\)