If a sequence of positive consecutive even integers has six terms, the

Let the 6 terms = \(n,n+2,n+4,n+6,n+8,n+10\)
Range = \(n+10 - n = 10\)
For Symmetric series = Median = Mean = \(n + 5\)
Product = \((n+5)10 = 50 + 10n\). Since, n can not be equal to 1 as it's even so the lowest value it can have is 2. Thus, for \(n = 2, 50 + 10(2) = 70. \)

Option (D)

Yeah took n+6 twice, also made calculation mistake ,my bad

I plugged numbers in and found 70 to work instantly, but a clear algebraic solution would be:

Let the first even number in the set be "2n", where 'n' is a POSITIVE integer (as defined by the question, 'n' can never be negative here).

Then our set of 6 consecutive even numbers is: 2n, 2n+2, 2n+4, 2n+6, 2n+8, 2n+10

So the median is the average of the middle two terms: (2n+4+2n+6)/2 = (4n+10)/2 = 2n+5
The range is simply 2n+10 - 2n = 10

The product of both (median*range) is 20n+50. Remember that 'n' has to be a positive integer as the sequence consists of positive even integers. This means that at the very least, the product of the median and range for these 6 positive consecutive integers has to be 70 (answer D). Please correct me if there is any error in logic here, I am not an expert.

The answer choices are all pretty low, so we can probably get away with just trying something to see if it works and adjusting as needed.
2, 4, 6, 8, 10, 12
Range = 10
Median = 7
Range * Median = 70
Answer choice D


But what sorts of things should we be alert to on questions like this that might be a bit tougher?
- We know the range of some set of consecutive even (or odd) integers is two times (the number of elements minus one). In this case, 2*(6-1) = 10.
- We know the median of some set of consecutive even integers is going to be even if there are an odd number of elements (since the median would be one of the elements) and odd if there are an even number of elements (since the median would be between the two in the middle). If you need practice, what would we know about the median of a set of consecutive odd integers?
- Taking those two things together, we'd know that we need an answer that is a multiple of 10 and that it must be the result of multiplying an odd number by 10. That would eliminate C and E.
- We could then notice that we won't have enough room to put half the elements before 3 or before 5, so A and B are out, and we are left with D.

let n be a positive even integer

n+n+2+n+4+n+6+n+8+n+10

Median= (n+4+n+6)/2= n+5
Range= n+10-n
Product=(n+5)*10

So 10n+50
Knowing n can’t be odd 20+50=70

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I have arrived at the answer in this manner, let me know whether is it the correct approach or not.

So basically, I chose some random set of even consecutive numbers like 2,4,6,8,10,12. by meeting the requirements of the question i have gotten the answer as 70.

Let the first number of the series be x and x>0. (since x is a positive even number)
Thus, the numbers will be x, x+2, x+4, x+6, x+8 and x+10.
Now, the median of the series = [(x+4) + (x+6)]/2 [Since the median of even term series is the average of the two middle terms]
= x+5

Range of series = Highest value term - lowest value term
= (x+10) - (x)
= 10

The product of range and median = 10*(x+5)
= 10x+50

Since x is an even positive integer, the answer will be greater than 50.
The only option possible is 70, i.e. when x=2.
Thus, the correct option is D.