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