Question

find the sum of all 3 digit numbers which are divisible by 3

Answer 1

Answer:

here you gooo

Step-by-step explanation:

165150

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Answer 2

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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)= 3897

n−1=299n=300

Now,

Sum of numbers in sequence= 2n

[2a+(n−1)d) = 2300

[2×102+(300−1)×3]=165150

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