Set N is a series of six consecutive odd integers. In set N, the lowes

Solution:

Since n^2 is the smallest value and 7n is the largest value, n^2 < 7n. Also, n can’t be negative; otherwise, the right hand side of n^2 < 7n would have been negative, whereas the left hand side is always nonnegative. So n must be positive (n can’t be 0 either since all integers in set N are odd). Therefore, when we divide both sides by n, we have n < 7. That is, n is 1, 3, or 5.

If n = 1, then n^2 = 1 would be the smallest value and we would have set N = {1, 3, 5, 7, 9, 11}. However, we see that, in this case, the largest value would be 11, which is not equal to 7n.

If n = 3, then n^2 = 9 would be the smallest value and we would have set N = {9, 11, 13, 15, 17, 19}. However, we see that, in this case, the largest value would be 19, which is not equal to 7n.

If n = 5, then n^2 = 25 would be the smallest value and we would have set N = {25, 27, 29, 31, 33, 35}. We see that, in this case, the largest value, 35, is indeed equal to 7n. Lastly, we see that the median of set N is (29 + 31)/2 = 60/2 = 30.

Alternate Solution:

Since there are six consecutive odd integers in set N and since n^2 is the smallest element of N, the largest element in N must be n^2 + 10. If we don’t see this, we can simply write N as N = {n^2, n^2 + 2, n^2 + 4, n^2 + 6, n^2 + 8, n^2 + 10}. We are also told that the largest element in N is equal to 7n, so we can write:

n^2 + 10 = 7n

n^2 - 7n + 10 = 0

(n - 2)(n - 5) = 0

n = 2 or n = 5

Since we are told that n^2 is odd, n cannot equal 2; thus, it must be true that n = 5. Hence, the set N can be expressed as N = {25, 27, 29, 31, 33, 35). The median of this set is (29 + 31)/2 = 30.

Answer: D

Odd consecutive integers will have even median hence the right ans choice will be either 10 or 30

If the median is 10 then it lies between 9 and 11 6th integer will be 15 which is not a multiple of 7 so the right and choice is 30 hence D is correct ans

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I may have accidentally gotten the right answer. Can someone please confirm if this method works as well. I saw that there is 6 consecutive integers, thus, I assume the median would have to be a multiple of 6. Is this approach correct?

Hello,

Great way to solve the sum! Kudos

This is an arithmetic progression. In AP mean = median. And,

mean = (first term + last term)/2

Also odd + odd = even
And, even/2 = even

This implies mean has to be even.

mean = (n^2 + 7n)/2

Try n=3. No option for this
n=5. Matches option D.