NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.13 Solutions : How many three digit numbers are divisible by 7?
Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : ”How many three digit numbers are divisible by 7?” of Exercise 5.2 Question No 13 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. 13 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 – 13 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. 13 :
How many three digit numbers are divisible by 7?
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
an = 994
n = ?
an = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
Alternative Method :
Three digit numbers which are divisible by 7 are 105, 112, 119, …. 994 .
These numbers form an AP with a = 105 and d = 7.
Let number of three-digit numbers divisible by 7 be n, an = 994
⇒ a + (n – 1) d = 994
⇒ 105 + (n – 1) × 7 = 994
⇒7(n – 1) = 889
⇒ n – 1 = 127
⇒ n = 128
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