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

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## Question

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

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 13

## 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 $$7.$$

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