12 Days of Christmas GMAT Competition - Day 7: j, 9, k, and m are in a

Let's visualise the two sequences:
j (+d) 9 (+d) k (+d) m
j (*a) k (*a) m

Interestingly, the difference between k and m is just once \(d\), but between j and k - it's twice as big, it's \(d\) two times.
This means that in the geometric progression is not growing - it seems to be diminishing. So, the ratio of the progression is definitely not higher than 1.
Let's calculate it:
\(k - j = 2d\\
m - k = d\)
Therefore, \(k - j = 2(m - k)\\
aj - j = 2(ak - k)\\
j(a-1) = 2k(a-1)\\
j(a-1) - 2k(a-1) = 0\\
(a-1)(j-2k) = 0\)
From here, we get two options for the common ratio: \(a = 1\) and \(a = 0.5\)
I know that theoretically, it looks odd if all the members of the progression are equal - but we have nothing stating that this is impossible.
Let's finalise these two scenarios and then look at the given conditions:

• \(a=1\), so \(j=k=m=9\)
• \(a=0.5\), so \(k = \frac{j}{2}\), which means \(2d = \frac{j}{2} \) and \(d=\frac{j}{4}\),
  so \(j = 9 + \frac{j}{4}\) and \( j = \frac{9}{3/4} = 12\)
  And thus, \(j = 12, k = 6, m =3\)

Now let's consider the conditions:

1. \(k+3 + 2m = 12 + 18 = 30 \): divisible by 5.
2. \(k+3 + 2m = 9 + 6 = 15 \): divisible by 5.

So, Condition 1 is not enough, it doesn't help us at all.


Without further talk, this simply tells us that all members of the sequence cannot be equal. This only leaves us one scenario: \(j*k=12*6=72\)
So, Condition 2 is enough by itself - Answer B.

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