**NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.16**** Solutions : ****Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.**

Hy Friends Welcome On **NCERT MATHS SOLUTIONS** !! Today we are going to solve the Question : **”****Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.****” **of** Exercise 5.2 Question No 16 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. 16 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 – 16** **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. 16 :**

**Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.**

**Class 10 Exercise 5.2 Question No. 16 Solutions**

*a*_{3}= 16*a* + (3 − 1) *d* = 16

*a* + 2*d* = 16 … **(i)**

**a _{7} − a_{5} = 12**

*a*+ (7 − 1)

*d*] − [

*a*+ (5 − 1)

*d*]= 12

(

*a*+ 6

*d*) − (

*a*+ 4

*d*) = 12

2

*d*= 12

*d*= 6

From equation

**(i)**, we get,

*a*+ 2 (6) = 16

*a*+ 12 = 16

*a*= 4

**Therefore, A.P. will be**

**4, 10, 16, 22, …**

## **OR**

**Solution:**

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