Question

How many three-digit numbers are divisible by 9 ?

Give solution in detail that how u get the terms such as first term , last term, etc

Answer 1

Answer:

The first three digit number is 100. The first three digit number divisible by 9 is 108.

This can be obtained by adding 9 to the largest 2 digit number divisible by 9. Largest 2 digit number divisible by 9 is 99. Hence adding 9 we get, 108 as the three digit number which is the first term. Last three digit number is 999 and eventually it is divisible by 9. So last term is 999.

So, a = 108, d = 9, l = 999, n = ?

We know that,

l = a + ( n - 1 ) d

=> 999 = 108 + ( n - 1 ) 9

=> 999 - 108 = ( n - 1 ) 9

=> 891 = ( n - 1 ) 9

=> 891 / 9 = ( n - 1 )

=> 99 = n - 1

=> n = 99 + 1 = 100 terms

Hence there are 100 three digit terms between 108 and 999 which are divisible by 9.

Answer 2

HEY THERE!!

Question:-

How many three-digit numbers are divisible by 9 ?

Method of Solution:-

3three-digit numbers are divisible by 9 ?

digits Number which are divisible by 9 given below in the form of Arithmetic Sequence or Progression;

108,117,126,....999

Above Equation is it in the form of Arithmetic Sequence or Progression;

Here,

First term=108

Common difference=117-108

Common difference=9

Last term=999

Now, Using the Formula of Arithmetic Sequence or Progression

Let it's Contain   'n'  ,Then,

Tn=999

=> a+(n-1)d

999=>108+(n-1)9

999=> 108+9n-9

999=> 99+9n

999-99=> 9n

900= 9n

•°• n=900/9

   n=100

Hence,(100),three-digit numbers are divisible by 9 .