**NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.8**** Solutions : ****An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.**

Hy Friends Welcome On **NCERT MATHS SOLUTIONS** !! Today we are going to solve the Question : **”****An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.****” **of** Exercise 5.2 Question No 8 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. 8 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 – 8** **Solutions** . Hope you like this post about Class 10th Maths Solution.

Here we would like to tell you that there are other websites on which the students demanded the Solution of NCERT MATHS Chapters, we had promised them that we will soon send you the questions of Class 9, And we completed our promise to them.We asked the good knowledgeable teachers of mathematics and asked them to solve the issues of CBSE’s mathematical questions and also said that the answers to our questions are such that the students have no difficulty in solving all those questions.

**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. 8 :**

**An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.**

**Class 10 Exercise 5.2 Question No. 8 Solutions**

Given that,

*a*_{3} = 12

*a*_{50} = 106

We know that,

**a _{n} = a + (n − 1) d**

*a*

_{3}=

*a*+ (3 − 1)

*d*

12 =

*a*+ 2

*d*…

**(i)**

Similarly,

*a*

_{50 }=

*a*+ (50 − 1)

*d*

106 =

*a*+ 49

*d*…

**(ii)**

On subtracting

**(i)**from

**(ii)**, we get

94 = 47

*d*

*d*= 2

From equation

**(i)**, we get

12 =

*a*+ 2 (2)

*a*= 12 − 4 = 8

*a*

_{29}=

*a*+ (29 − 1)

*d*

*a*

_{29}= 8 + (28)2

*a*

_{29}= 8 + 56 = 64

Therefore,

**29**

^{th}term is 64.## **OR**

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