**NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.11**** Solutions : ****Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?**

Hy Friends Welcome On **NCERT MATHS SOLUTIONS** !! Today we are going to solve the Question : **”****Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?****” **of** Exercise 5.2 Question No 11 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. 11 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 – 11** **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. 11 :**

**Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?**

**Class 10 Exercise 5.2 Question No. 11 Solutions**

**Given A.P. is 3, 15, 27, 39, …**

*a *= 3

*d* = *a*_{2} − *a*_{1} = 15 − 3 = 12

*a*_{54} = *a* + (54 − 1) *d*

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to find the term of this A.P. which is 771.

**Let n ^{th} term be 771.**

**a**

_{n}= a + (n − 1) d771 = 3 + (

*n*− 1) 12

768 = (

*n*− 1) 12

(

*n*− 1) = 64

*n*= 65

Therefore, 65

^{th}term was 132 more than 54

^{th}term.

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**Solution:**

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