Answer:
[tex]11, 13, 15[/tex]
B) 6x + 16 = 82
Note: I just kept my wrong approach.
Step-by-step explanation:
Let first define an odd integer.
Considering [tex]x\in\mathbb{Z}: 2k+1[/tex], [tex]x[/tex] is an odd integer.
Now, three consecutive odd integers are [tex]a_1 = 2k+1[/tex] , [tex]a_2 = 2k+3[/tex] and [tex]a_3 = 2k+5[/tex]
The sum of the first, two times the second and three times the third can be written as
[tex]a_1 + 2a_2 + 3a_3 = 2k+1 + 2(2k+3)+3(2k+5)[/tex]
Expanding and simplifying
[tex]2k+1 + 2(2k+3)+3(2k+5) = 2k+1 +4k+6+6k+15 = 12k+22[/tex]
This is what I have done initially, but considering [tex]x[/tex] to be an odd integer, three consecutive integers can be written as
[tex]x, x+2, x+4[/tex] such that [tex]x\in\mathbb{Z}: 2k+1[/tex]
Therefore,
[tex]x+2(x+2)+3(x+4)=82 \implies x+2x+4+3x+12 = 82 \implies 6x+16 = 82[/tex]
and [tex]6x+16 = 82 \implies 6x = 66 \implies x = 11[/tex]
The three consecutive odd integers such that the sum of the first, two times the second and three times the third is 82 are [tex]11, 13, 15[/tex]
