Maths!
posted by Ariana .
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!

300 = 100 + 40*5
So, set things up as an arithmetic sequence, and find that
S41 = 41/2 (100+300) = 8200 
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 nth term this arithmetic progression an = 300
Use formula for nth 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
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