Question

Find the sum of all multiples of 5 between 100 and 300 inclusive.

Ans: 8200

I do not know how to do this. Really do appreciate step by step workings and explanation.
Thanks!

Answer 1

300 = 100 + 40*5
So, set things up as an arithmetic sequence, and find that

S41 = 41/2 (100+300) = 8200

Answer 2

100 , 105, 110, ... 290 , 295 ,300


That is arithmetic progression.

The initial term of this arithmetic progression is a1 = 100

The common difference of successive members is d = 5

The n-th term this arithmetic progression an = 300


Use formula for n-th term of the sequence :

an = a1 + ( n - 1 ) * d


In this case :

a1 = 100

an = 300

d = 5


an = a1 + ( n - 1 ) * d

300 = 100 + ( n - 1 ) * 5 Divide both sides by 5

60 = 20 + ( n - 1 )

60 = 20 + n - 1

60 = 19 + n Subtract 19 to both sides

60 - 19 = 19 + n - 19

41 = n

n = 41


The sum of the members of a arithmetic progression is :

Sn = ( n / 2 ) * ( a1 + an ) = n * ( a1 + an ) / 2

Sn = 41 * ( 100 + 300 ) / 2

Sn = 41 * 400 / 2

Sn = 16,400 / 2

Sn = 8,200