**NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.13**** Solutions : ****How many three digit numbers are divisible by 7?**

Hy Friends Welcome On **NCERT MATHS SOLUTIONS** !! Today we are going to solve the Question : **”****How many three digit numbers are divisible by 7?****” **of** Exercise 5.2 Question No 13 Solutions (Arithmetic Progressions)** of Class 10th, which will prove to be very helpful for you.

**CBSE/NCERT MATHS Class 10 Ex-5.2 Q No. 13 Solutions**

If you are a student of CBSE, today we are going to give you the **CBSE / NCERT** **Chapter : Arithmetic Progressions ** **Exercise 5.2 Question No – 13** **Solutions** . Hope you like this post about Class 10th Maths Solution.

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**Arithmetic Progressions CHAPTER : 5**

n this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term.

Suppose, a_{1}, a_{2}, a_{3}, ……………., a_{n} is an AP, then;_{ }the **common difference “ d ”** can be obtained as; In this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term.

**a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d**

**Exercise- 5.2 Question No. 13 :**

**How many three digit numbers are divisible by 7?**

**Class 10 Exercise 5.2 Question No. 13 Solutions**

**First three-digit number that is divisible by 7 = 105**

**Next number = 105 + 7 = 112**

Therefore, 105, 112, 119, …

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.

**The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.**

The series is as follows.

**105, 112, 119, …, 994**

Let 994 be the *n*th term of this A.P.

*a* = 105

*d* = 7

*a*_{n} = 994

*n* = ?

**a _{n} = a + (n − 1) d**

994 = 105 + (

*n*− 1) 7

889 = (

*n*− 1) 7

(

*n*− 1) = 127

*n*= 128

Therefore,

**128 three-digit numbers are divisible by 7.**

**Alternative Method :**

**Three digit numbers which are divisible by 7 are 105, 112, 119, …. 994 .**

These numbers form an AP with *a* = 105 and *d* = 7.

Let number of three-digit numbers divisible by 7 be *n*, *a*_{n} = 994

**⇒ a + (n – 1) d = 994**

⇒ 105 + (*n* – 1) × 7 = 994

⇒7(*n* – 1) = 889

⇒ *n* – 1 = 127

⇒ *n* = 128

## **OR**

**Solution:**

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