Question

how many three digit numbers are there divisible by 3 find their sum.​

Answer 1

Answer:

165150

Step-by-step explanation:

Smallest 3 digit number divisible by 3 a=102

Largest 3 digit number divisible by 3 a

n

=999

Common difference d=3

Let the total number of 3 digit number be n

Therefore,

a

n

=a+(n−1)d

substituting the values

999=102+(n−1)×3

999−102=(n−1)×3

897=(n−1)×3

(n−1)=

3

897

n−1=299

n=300

Now,

Sum of numbers in sequence=

2

n

[2a+(n−1)d]

=

2

300

[2×102+(300−1)×3]

=165150

Therefore sum of all 3 digit number which are divisible by 3 is 165150.

Answer 2

Answer:

Smallest 3 digit number divisible by 3 a=102

Largest 3 digit number divisible by 3 a

n

=999

Common difference d=3

Let the total number of 3 digit number be n

Therefore,a

n

=a+(n−1)d

substituting the values

999=102+(n−1)×3

999−102=(n−1)×3

897=(n−1)×3

(n−1)=

3

897

n−1=299

n=300

Now,

Sum of numbers in sequence=

2

n

[2a+(n−1)d]

=

2

300

[2×102+(300−1)×3]

=165150

Therefore sum of all 3 digit number which are divisible by 3 is 165150.