Find the sum of the multiples of 6 between 115 and 245.
first multiple of 6 between 115 and 245 is 120
last multiple of 6 between 115 and 245 is 240
so you have an arithmetic series
120 + 126 + .. + 234 + 240
a = 120 , d = 6 , n = ?
finding how many terms ....
term(n) = 240
a + (n-1)d = 240
120 + (n-1)(6) = 240
6n - 6 = 120
6n = 126
n = 21
Sum(n) = (n/2)(first + last)
= (21/2)(120+240) = 3780