(02.02 lc) choose the equation that could be used to find three consecutive integers whose sum is 36. (5 points) n + (n + 2) + (n + 4) = 36 n + (n + 1) + (n + 3) = 36 n + (n + 1) + (n + 2) = 36 n + (n − 1) + (n − 3) = 36

n + (n + 1) + (n + 2) = 36 Let's first create the equation ourselves and then see if it fits any of the available options. We want 3 consecutive integers that add up to 36. Let's use the value "n" to represent the first number in the sequence. n Now the next number will be n+1 and the third will be n+2. And since we want the sum, we get n + n + 1 + n + 2 = ? And finally, we want the sum to be 36, so our final equation is n + n + 1 + n + 2 = 36 With this equation in mind, let's look at the available choices. n + (n + 2) + (n + 4) = 36 Well, it has the 36 OK, but it has +2 and +4 where we have +1 and +2. So it won't fit and it's wrong. n + (n + 1) + (n + 3) = 36 Closer. Got the 36 and the +1. But it has +3 instead of +2, so not a fit either. n + (n + 1) + (n + 2) = 36 This looks good. Has a couple of extra parenthesis, but they don't affect the final answer and it has the +1, +2 and the 36. This one is correct. n + (n â’ 1) + (n â’ 3) = 36 Hmm. The n - 1 has a possibility. Perhaps they want n to be the middle of the 3 numbers to add. If that's the case, then the other number should be n+1 which it isn't. Maybe they want the n to be the last number. If that's the case, then the third number should be n-2. But that doesn't work either. So this equation won't work. So of the 4 choices, the only answer that works is "n + (n + 1) + (n + 2) = 36"