akijuneja wrote:
The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?
(1) The ninth term in this sequence is 81.
(2) The fifth term in this sequence is 1
I'm going to try to explain this one in a less 'mathematical' way, as an alternative for those of us who aren't super enthusiastic about algebra. Sequences are also pretty easy to translate into plain English, and that's how I prefer to think about them.
The question stem says that we're going to build a sequence by starting at some number, then multiplying by the same value over and over until we have 9 terms. For example, this would fit the rules:
1, 3, 9, 27, 81, 243, ... etc.
This would also fit the rules:
4, 2, 1, 1/2, 1/4, 1/8, 1/16, 1/32, ... etc.
This would
not fit the rules, since k can't equal 1 (we can't just multiply by 1 over and over):
3, 3, 3, 3, 3, 3, ...This also wouldn't work, since we have to
multiply by the same value instead of adding the same value:
1, 2, 3, 4, 5, ...So that's what the question is telling us. Notice that it's
not telling us what the first number in the sequence is, so that could be literally anything. The first number could even be negative! (We just can't multiply by a negative number to get the next term.)
Finally, the question itself asks how many of the first 9 terms are greater than 1. I'm not sure how to figure that out mathematically, to be honest, so I'm going to focus on case testing.
Now, the statements.
Statement 1 The ninth term is 81.
_ _ _ _ _ _ _ _ 81
I know that I started with some value (which I don't know) and then multiplied by the same number over and over, until I hit 81 after multiplying 8 times.
In order to go
backwards, I'll have to start with the 81, and divide by the same number, over and over.
Suppose that k = 3. Build the sequence backwards:
_ _ _ _ _ _ _ 27 81
_ _ _ _ _ _ 9 27 81
_ _ _ _ _ 3 9 27 81
_ _ _ _ 1 3 9 27 81
The other four terms will be less than 1. So the answer to the question is "4 terms".
Now let's try a really crazy case. Suppose that k = 81. That is, we'll be dividing by 81 over and over.
_ _ _ _ _ _ _ 1 81
_ _ _ _ _ _ 1/81 1 81
1/81 is less than 1, and so are the other 6 terms at the beginning. The answer the question is "7 terms".
This statement is
not sufficient.
Statement 2: The fifth term of the sequence is 1. So, it looks like this:
_ _ _ _ 1 _ _ _ _
We'll be going in both directions now. To go to the left, we'll divide by k. To go to the right, we'll multiply by k.
Let's say that k = 2.
_ _ 1/4 1/2 1 2 4 _ _
Interesting. The four to the left are less than 1. The answer to the question is "4".
Let's say that k = 1/2.
_ _ 4 2 1 1/2 1/4 _ _
Okay, now the four to the right are less than 1. The answer to the question is "4".
Will that always be true? Yeah - if k is a fraction, then the four numbers to the right will all be smaller than 1. If k isn't a fraction, the four numbers to the left will all be smaller than 1. That's because we're either dividing 1 by a number repeatedly (making it smaller), or multiplying 1 by a number repeatedly (making it bigger).
That's why this statement is sufficient,
even if k is a fraction. _________________