**NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.17**** Solutions : ****Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.**

Hy Friends Welcome On **NCERT MATHS SOLUTIONS** !! Today we are going to solve the Question : **”****Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.****” **of** Exercise 5.2 Question No 17 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. 17 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 – 17** **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. 17 :**

**Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.**

**Class 10 Exercise 5.2 Question No. 17 Solutions**

**Given A.P. is**

**3, 8, 13, …, 253**

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 5

For this A.P.,

*a* = 253

*d* = 248 − 253 = −5

*n *= 20

*a*_{20} = *a* + (20 − 1) *d*

*a*_{20} = 253 + (19) (−5)

*a*_{20} = 253 − 95

*a* = 158

Therefore, **20 ^{th} term from the last term is 158.**

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**Solution:**

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