Determine the total sum of all three-digit numbers that are

Solution

A number which is divisible by both 2 and 7 is a number which is divisible by 14 (LCM of 2 and 7)
Smallest 3-digit number divisible by 14 = 112
Largest 3-digit number divisible by 14 = 994
Therefore, numbers will be 112, 126,......, 994 which are in arithmetic progression having
First term (a) = 112, common difference (d) = 14 and last term (l) = 994
Last term of a series, l = a + (n - 1) X d
Where, 'a' is the first term of the series, 'n' is number of terms, and 'd' is the common difference of the series.
Or, 994 = 112 + (n - 1) X 14
Or, 'n' - 1 = (994 - 112) ÷ 14
So, 'n' = 63 + 1 = 64
Sum of a series in an arithmetic progression, Sn = (n/2) X {2a + (n - 1) X d}
Where, 'a' is the first term of the series, 'n' is number of terms, and 'd' is the common difference of the series.
= (64/2) X {2 X 112 + (64 - 1) X 14}
= 32 X (224 + 882) = 32 X 1106 = 35,392