Question

What is the sum of all 3 digit number which are divisible by 3 but not 6
Class 10

Answer 1

First number = 105
Last number = 999
The common difference is 6.
999=105+(n-1)6
6n-6=894
6n=900
n=150
Sum=n/2(a+l)
=150/2(105+999)
=75*1104
=82800

Answer 2

The first three-digit number divisible by 3 but not divisible by 6 is a= 105.

Next term is 111, 117, 123, 129......999

The last three-digit number divisible by 6 but not divisible by 6 is tn = 999.

The common difference is d = 6.


We know that tn = a + (n - 1) * d

                       999 = 105 + (n - 1) * 6

                       999 = 105 + 6n - 6

                       999 = 99 + 6n

                       999 - 99 = 6n

                       990 = 6n

                       150 = n.


Now,

The sum of n terms of an AP = n/2(a + l)

                                                 = 150/2(105 + 999)

                                                 = 75 * 1104

                                                 = 82800.


Hence the sum of all three-digit numbers divisible by 3 but not by 6 = 82800.


Hope this helps!