Maths!

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!

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asked by Ariana
  1. 300 = 100 + 40*5
    So, set things up as an arithmetic sequence, and find that

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

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    posted by Steve
  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

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    posted by Bosnian

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