NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.14 Solutions : How many multiples of 4 lie between 10 and 250?

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NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.14 Solutions : How many multiples of 4 lie between 10 and 250?

Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : How many multiples of 4 lie between 10 and 250?of Exercise 5.2 Question No 14 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. 14 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 – 14 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, a1, a2, a3, ……………., an 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. 14 :

How many multiples of 4 lie between 10 and 250?

Class 10 Exercise 5.2 Question No. 14 Solutions

First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
a = 12
d = 4
an = 248
an = a + (n – 1) d
248 = 12 + (n – 1) × 4
236/4 = n – 1
59  = n – 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.

Alternative Method :

Multiples of 4 lies between 10 and 250 are 12, 16, 20, …., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be nan = 248
⇒ a + (n – 1) d = 248
⇒ 12 + (n – 1) × 4 = 248
⇒4(n – 1) = 248
⇒ n – 1 = 59
⇒ n = 60

                                          OR

Solution:
Chapter 5 Maths Class 10 NCERT Solutions Arithmetic Progression Ex 5.2 Q14

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