Find the sum of all multiples of 5 lie between 19 to 556.
the first one would be 20 and the last one 555
so treat it like an AP with a = 20 and d = 5
how many terms are there?
t(n) = a + (n-1)d
555 = 20 + 5(n-1)
n-1 = 107 , so n = 108
We have 108 terms.
Use the formula
sum(n) = (n/2)(first + last)
= ....
let me know what you got.