SOLUTION: find the sum of the multiples of 3 from 3 to 99 inclusive

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Question 427328: find the sum of the multiples of 3 from 3 to 99 inclusive
Found 2 solutions by Theo, Edwin McCravy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the multiples of 3 are 3,6,9,12,15,18,21,24,27,30,......... all the way up to 99.

this is an arithmetic series with a common difference of 3.

the formula for the sum of an arithmetic series is:

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

Sn = the sum of the n terms in the sequence.
a1 = the first term in the sequence.
an = the nth term in the sequence.

our first term is 3 and our last term is 99.

the common difference is 3.

to find the number of terms, we take the last term and divide it by 3 to get 33.

there are 33 terms in the sequence, so n = 33

our formula becomes Sn = (33/2 * (3 + 99) which becomes (33/2) * 102 which becomes 33 * 51 which becomes 1683.

to see how this works, we can use smaller numbers.

assume the first term is 3 and the last term is 9.

the common difference is 3.

the number of terms is 9/3 = 3

the formula states that Sn = (3/2) * (3 + 9) = (3/2) * (12) = 3*6 = 18

if we sum up the terms in the sequence we should equal 18.

3 + 6 + 9 = 9 + 9 = 18.

the formula works.

same thing happens with the larger numbers only it's much harder to show you because there's so many more numbers to add up.





Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
find the sum of the multiples of 3 from 3 to 99 inclusive

Two ways:  with and without formulas:

1. With formulas:

an = a1 + (n - 1)d
Sn = (n/2)(a1 + an)

a1 = 3
 d = 3
 
an = a1 + (n - 1)·d
99 =  3 + (n - 1)·3
99 =  3 + 3(n - 1)
96 = 3(n - 1)
32 = n - 1
33 = n

 Sn =  (n/2)(a1 + an)
S33 = (33/2)(3 + 99)
S33 = (33/2)(102)
S33 = 1683 

2:  Without formulas:

 S =    3 +  6 +  9 + ... + 97 + 98 + 99
 S = 3( 1 +  2 +  3 + ... + 31 + 32 + 33)
 S = 3(33 + 32 + 31 + ... +  3 +  2 +  1)
2S = 3(34 + 34 + 34 + ... + 34 + 34 + 34)
2S = 3(34·33)
2S = 3366
 S = 1683

Edwin
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