An increasing integer sequence with 5 elements has 13 as its third ele

Question

An increasing integer sequence with 5 elements has 13 as its third element. If the mean, median, mode, and range of the sequence are equal, then which of the following must be true?

I. The least element in the sequence is 7. 
II. The greatest element in the sequence is 20. 
III. No sequence can be formed based on the conditions given.

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) None of the above.

Abhishekgmat87 · Answer

Median is 13. So the 3rd number is 13. Mean is 13. So, the sum of the 5 numbers will be 65
Now range is 13. Difference of 5th and 1st term is 13. Their sum will be odd. (7,20) or (8,21)
This imply that the sum of the 2nd , 3rd and 4th term will be even. Odd +  even  = Odd

So all the 3 can't be 13. As mode is also 13. There are 2 number 13 in the sequence.

Let's check the statements -
I. The least element in the sequence is 7.
Let us take smallest number as 6. Largest number will be 20. 
Sum of 2nd to 4th numbers = 40
There are 2 13's . So one number will be 14. The series will be 6,13,13,14,19. This imply that the smallest number is 7 is not true

II. The greatest element in the sequence is 20. 
Let's take greatest number as 21. The least number will be 8. 
Sum of 2nd to 4th numbers = 36
There are 2 13's . So one number will be 10. The series will be 8,10,13,13,21. Statement II is not true

Statement III is also not true as the sequence are possible as we checked in 1st 2 statements

Option E

riddhisiddhi · Answer

It says increasing intergee sequence.
I suppose not means that all the numbers will be innoncreasing order
So how two 13 are possible
I selected C option , considering this

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Abhishekgmat87 · Answer

Since mode is 13, we have to assume that 13 appear more than once in the sequence. Increasing integer sequence is given to tell us that the numbers are arranged in ascending order. The 3rd term will be mean and median.

I might be wrong.

va789 · Answer

I have the same question. As to the argument that "mode is 13 so 13 must appear at least twice", I'm confused as a sequence can have more than one mode.

Bunuel   VeritasKarishma   BrentGMATPrepNow   chetan2u  other experts- Please help clear this doubt.

Additionally, are questions framed in such ambiguous language on the actual GMAT?

Thanks for your help!

BrentGMATPrepNow · Answer

I don't believe the GMAT would use the phrase  increasing integer sequence  to describe a sequence in which two values are the same. 
If we're told the sequence is  increasing , then we can conclude that term1 < term2 < term3 < term4....etc

KarishmaB · Answer

GMAT is likely to give that the numbers are arranged in ascending order. 
Also, a sequence in which every number appears exactly once has no mode. So for the mode to be 13, 13 has to appear at least twice. Also, if a sequence has more than one mode, we cannot correctly say that the mode of the sequence is 13. We need to say that the modes of the sequences are 13 and say 15. 
But, we can say that 13 is a mode of the sequence (implying that there could be other modes too).

We need mean = median = mode = range = 13
Since median is 13, the 3rd integer is 13.
Since 13 is the mode too, 2nd or 4th integer needs to be 13 too.
For range to be 13, Fifth - First = 13
For mean to be 13, whatever is the deficit from 13, the same should be the excess from 13.

The easiest, most uniform case would be __ 13, 13, 13 __ such that first and fifth numbers are each 6.5 away from 12. But then the first and fifth numbers will not be integers. So middle 3 numbers cannot be 13. 13 must appear only twice.

So say fifth number is 7 away from 13 and first number is 6 away from 13. 
7 __ 13, 13, 20
The deficit is 6 and excess 7 so the list becomes 
7, 12, 13, 13, 20

But there are other cases possible too. 
8, 10, 13, 13, 21
etc.

Hence, none of the statements are correct. 
Answer (E)

axbycz37 · Answer

VeritasKarishma   Bunuel 
Here's an analysis:

Inference 1  - The given info states that the mean, median, mode, and range are equal and that this is an increasing integer sequence. If the mean and median of any increasing integer sequence are equal, then the sequence has to be in an AP.
 Inference 2  - An AP can never have a mode unless all the value of the AP are the same number because otherwise all the the value in the AP will be different and thus, the sequence will have no mode.
 Conclusion  - All the numbers in this sequence are the same - 13 (as we are told that the 3rd element is 13).

There seems to be an inherent contradiction in this question - If this is an "increasing integer" sequence, then there cannot be a mode.
Let's look at our options:

Answer - Ideally, it should be C because there cannot be an increasing integer sequence where mean and median are equal and there exists a mode.
What do you think?

KarishmaB · Answer

This is not correct. Consider some non AP sequences:
1, 4, 6, 8, 11 
or 
1, 3, 6, 7, 13
etc
Mean = Median = 6

Nirmesh83 · Answer

Therefore , if a number is in AP then Mean =mode but vice-versa is not true .

elslyknight · Answer

For E to be the answer, the Question stem is assuming that it is an increasing sequence means a non-decreasing sequence, where some numbers can be same (hence, we have a mode) - But that is largely not how one would internpret that stem. One would interpret the stem to be a distinct 5 numbers. Hence C is more appropriate.

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An increasing integer sequence with 5 elements has 13 as its third ele

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