algebra and trigonometry

Question

Find the sum of all the positive integers less than 100 that are multiples of 3.

Answer 1

There are 100/3 = 33 positive numbers < 100 that are multiples of 3 or the sequence 3,6,9,12,   ...  99. Dividing by 3 we have 1,2,3,4  ..  33. Then the answer is 3*(1,2,3,4 .. 33) Using Gauss's formula the sum of 1,2,3,4 .. 33 is n(n+1)/2 = (33*34)/2 = 561. Then multiplying by 3 gives the answer 1683.

Answer 2

100/3=33 terms

3,6,9,12,...90,93,96,99

there are different solutions

use the formula for an arithmetic sequence

S(33)=(33/2)(2*3+[33-1]3) where 3 is the first term and 3 is the common difference also

S(33)=(33/2)(6+32*3)

S(33)=(33/2)(6+96)

S(33)=(33/2(102)

S(33)=33*51

S(33)=1683

another solution without using the formula...

there are 33 terms

the 17th term is the middle term with 16 on the left and 16 on the right

3*16=48, so 51 is the middle term

add the first term and the last term to get 102

add the second term and the next to last term to get 102

continue this and you always get a sum of 102

there are 32 numbers plus the middle number of 51 for a total of 33 terms

there will be 32/2=16 sums of 102 plus the middle term of 51

16*102=1632, 1632+51=1683 again