The Sum Of All 3 Digit Numbers Divisible By 3 Is?

1.	The Sum Of All 3 Digit Numbers Divisible By 3 Is?
Answer» All 3 digit numbers DIVISIBLE by 3 are : 102, 105, 108, 111, ..., 999. This is an A.P. with first element 'a' as 102 and difference 'd' as 3. Let it contains n terms. Then, 102 + (n - 1) x3 = 999 102 + 3n-3 = 999 3n = 900 or n = 300 Sum of AP = n/2 [2a + (n-1)d] REQUIRED sum = 300/2[2102 + 2993] = 165150. All 3 digit numbers divisible by 3 are : 102, 105, 108, 111, ..., 999. This is an A.P. with first element 'a' as 102 and difference 'd' as 3. Let it contains n terms. Then, 102 + (n - 1) x3 = 999 102 + 3n-3 = 999 3n = 900 or n = 300 Sum of AP = n/2 [2a + (n-1)d] Required sum = 300/2[2102 + 2993] = 165150.

Answer»

All 3 digit numbers DIVISIBLE by 3 are :

102, 105, 108, 111, ..., 999.

This is an A.P. with first element 'a' as

102 and difference 'd' as 3.

Let it contains n terms. Then,

102 + (n - 1) x3 = 999

102 + 3n-3 = 999

3n = 900 or n = 300

Sum of AP = n/2 [2*a + (n-1)*d]

REQUIRED sum = 300/2[2*102 + 299*3] = 165150.

All 3 digit numbers divisible by 3 are :

102, 105, 108, 111, ..., 999.

This is an A.P. with first element 'a' as

102 and difference 'd' as 3.

Let it contains n terms. Then,

102 + (n - 1) x3 = 999

102 + 3n-3 = 999

3n = 900 or n = 300

Sum of AP = n/2 [2*a + (n-1)*d]

Required sum = 300/2[2*102 + 299*3] = 165150.

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