Question

16. Find the number of all three-digit natural numbers, which are divisible by 9. (CBSE 2013)​

Answer 1

Answer:

Thank me later

Step-by-step explanation:

Since the numbers are to be 3 Digit they will be more than 99 and less than 1000

The first no. is

108 then 117 and last is 999 (Can check dividing them)

So basically it is an AP with first term 108 and last term 999 and common difference 9

now how many terms in AP using Nth term=a+(n-1)d

so 999=108+(n-1)9

n=100

Sum of AP=n/2(a+l) where a is first term and l is last

So

100/2(108+999)=55350

Answer 2

Answer:

108, 117, 126, ……, 999 is 100. Therefore, the number of three-digit natural numbers divisible by 9 is 100.