Question

Find three consecutive odd integers such that the sum of the​ first, two times the second and three times the third is 82.

Answer 1

I think the answer is (A)

Answer 2

Answer:

$11, 13, 15$

B) 6x + 16 = 82

Note: I just kept my wrong approach.

Step-by-step explanation:

Let first define an odd integer.

Considering $x\in\mathbb{Z}: 2k+1$, $x$ is an odd integer.

Now, three consecutive odd integers are $a_1 = 2k+1$ , $a_2 = 2k+3$ and $a_3 = 2k+5$

The sum of the​ first, two times the second and three times the third can be written as

$a_1 + 2a_2 + 3a_3 = 2k+1 + 2(2k+3)+3(2k+5)$

Expanding and simplifying

$2k+1 + 2(2k+3)+3(2k+5) = 2k+1 +4k+6+6k+15 = 12k+22$

This is what I have done initially, but considering $x$ to be an odd integer, three consecutive integers can be written as

$x, x+2, x+4$ such that $x\in\mathbb{Z}: 2k+1$

Therefore,

$x+2(x+2)+3(x+4)=82 \implies x+2x+4+3x+12 = 82 \implies 6x+16 = 82$

and $6x+16 = 82 \implies 6x = 66 \implies x = 11$

The three consecutive odd integers such that the sum of the​ first, two times the second and three times the third is 82 are $11, 13, 15$