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# The next number in a certain sequence is defined by multiply

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Joined: 24 Jan 2017
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Re: The next number in a certain sequence is defined by multiply [#permalink]

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06 Jun 2017, 05:35
Bunuel 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?

Given that:
$$a_2=a_1*k$$
$$a_3=a_1*k^2$$
$$a_4=a_1*k^3$$
...
$$a_n=a_1*k^{n-1}$$

Also given that k>0 and n=9.

(1) The ninth term in this sequence is 81 --> $$a_9=a_1*k^8=81$$. If $$a_1=1$$, then all but $$a_1$$ will be greater than 1, but if $$a_1=2$$, then all will be greater than 1. Not sufficient.

(2) The fifth term in this sequence is 1 --> $$a_5=a_1*k^4=1$$. Now, if $$a_1<1$$, then $$k>1$$, and all terms from $$a_5$$ ($$a_6$$, $$a_7$$, $$a_8$$, and $$a_9$$), so 4 terms will be greater than 1 AND if $$a_1>1$$, then $$k<1$$, and all terms till $$a_5$$ ($$a_2$$, $$a_3$$ and $$a_4$$), so again 4 terms will be greater than 1. Sufficient.

I am not sure how statement 2 is sufficient if k is a fraction.?

Hi,

The reasoning of Bunuel about statement 2 is still correct considering k is a fraction. In fact, Bunuel did not eliminate this consideration (that k is a fraction).
- If k > 1 (no matter whether k is a fraction or not), then all 4 terms to the right of 5th term will be >1. It also means that all 4 terms to the left hand of 5th term will be <1.
Ex: k=3/2, then we have the following sequence: 16/81; 8/27; 4/9; 2/3; 1; 3/2; 9/4; 27/8; 81/16 . You can see that 4 terms to the right of 5th term are 3/2; 9/4; 27/8; 81/16 => All these 4 terms are >1

- Similarly, If k < 1 (no matter whether k is a fraction or not), then all 4 terms to the right of 5th term will be <1. It also means that all 4 terms to the left hand of 5th term will be >1.

If you still have concern(s), just indicate why you think statement 2 is insufficient if k is fraction, then I will response accordingly.
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The next number in a certain sequence is defined by multiply [#permalink]

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23 Jul 2017, 04:08
Bunuel 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?

Given that:
$$a_2=a_1*k$$
$$a_3=a_1*k^2$$
$$a_4=a_1*k^3$$
...
$$a_n=a_1*k^{n-1}$$

Also given that k>0 and n=9.

(1) The ninth term in this sequence is 81 --> $$a_9=a_1*k^8=81$$. If $$a_1=1$$, then all but $$a_1$$ will be greater than 1, but if $$a_1=2$$, then all will be greater than 1. Not sufficient.

(2) The fifth term in this sequence is 1 --> $$a_5=a_1*k^4=1$$. Now, if $$a_1<1$$, then $$k>1$$, and all terms from $$a_5$$ ($$a_6$$, $$a_7$$, $$a_8$$, and $$a_9$$), so 4 terms will be greater than 1 AND if $$a_1>1$$, then $$k<1$$, and all terms till $$a_5$$ ($$a_2$$, $$a_3$$ and $$a_4$$), so again 4 terms will be greater than 1. Sufficient.

Hi Bunuel, I don't quite understand the explanation for statement 2. You mention " if a1>1, then k<1, and all terms till a5(a2, a3 and a4), so again 4 terms will be greater than 1. Sufficient."
Why would K be <1? In the question it is clearly stated that K is a positive constant and does not equal 1. Thus, K will always be more than 1.

Please correct me if I am wrong. Thanks!
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Posts: 44647
Re: The next number in a certain sequence is defined by multiply [#permalink]

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23 Jul 2017, 05:56
roastedchips wrote:
Bunuel 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?

Given that:
$$a_2=a_1*k$$
$$a_3=a_1*k^2$$
$$a_4=a_1*k^3$$
...
$$a_n=a_1*k^{n-1}$$

Also given that k>0 and n=9.

(1) The ninth term in this sequence is 81 --> $$a_9=a_1*k^8=81$$. If $$a_1=1$$, then all but $$a_1$$ will be greater than 1, but if $$a_1=2$$, then all will be greater than 1. Not sufficient.

(2) The fifth term in this sequence is 1 --> $$a_5=a_1*k^4=1$$. Now, if $$a_1<1$$, then $$k>1$$, and all terms from $$a_5$$ ($$a_6$$, $$a_7$$, $$a_8$$, and $$a_9$$), so 4 terms will be greater than 1 AND if $$a_1>1$$, then $$k<1$$, and all terms till $$a_5$$ ($$a_2$$, $$a_3$$ and $$a_4$$), so again 4 terms will be greater than 1. Sufficient.

Hi Bunuel, I don't quite understand the explanation for statement 2. You mention " if a1>1, then k<1, and all terms till a5(a2, a3 and a4), so again 4 terms will be greater than 1. Sufficient."
Why would K be <1? In the question it is clearly stated that K is a positive constant and does not equal 1. Thus, K will always be more than 1.

Please correct me if I am wrong. Thanks!

We are told that k is positive constant and k ≠ 1. So, 0 < k < 1 or k > 1. Nowhere we are told that k is an integer.
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Re: The next number in a certain sequence is defined by multiply [#permalink]

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23 Jul 2017, 15:48
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Expert's post
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.
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Re: The next number in a certain sequence is defined by multiply [#permalink]

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08 Dec 2017, 12:59
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

We are given that the next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1, and we need to determine how many of the first nine terms in this sequence are greater than 1.

Statement One Alone:

The ninth term in this sequence is 81.

Without knowing any other terms in the sequence or the value of the positive constant k we are multiplying, we can’t determine the number of terms in the sequence that are greater than 1. For example, if k = 3, then by going backward we have:

9th term = 81, 8th term = 27, 7th term = 9, 6th term = 3, 5th term = 1, 4th term = 1/3, and so on.

In this case, we have 4 terms that are greater than 1.

However, if k = 9, then by going backward again we have:

9th term = 81, 8th term = 9, 7th term = 1, 6th term = 1/9, and so on.

In this case we have only 2 terms that are greater than 1. Therefore, statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

The fifth term in this sequence is 1.

Since there are 9 total terms, we see that the 5th term is the middle term in our sequence. In other words there are 4 terms below the 5th term and 4 terms above the 5th term. Since we know that k is positive, it’s either a positive proper fraction or a number greater than 1. Thus, regardless of whether k is a positive proper fraction or a number greater than 1, there will be 4 numbers above 1 and 4 numbers below 1.

For instance, if k = 1/2, then the 1st to the 4th terms, inclusive, are greater than 1, and if k = 2, then the 6th to the 9th terms, inclusive, are greater than 1. Either way, there are 4 terms greater than 1. Statement two alone is sufficient to answer the question.

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Re: The next number in a certain sequence is defined by multiply [#permalink]

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22 Apr 2018, 19:03
Bunuel 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?

Given that:
$$a_2=a_1*k$$
$$a_3=a_1*k^2$$
$$a_4=a_1*k^3$$
...
$$a_n=a_1*k^{n-1}$$

Also given that k>0 and n=9.

(1) The ninth term in this sequence is 81 --> $$a_9=a_1*k^8=81$$. If $$a_1=1$$, then all but $$a_1$$ will be greater than 1, but if $$a_1=2$$, then all will be greater than 1. Not sufficient.

(2) The fifth term in this sequence is 1 --> $$a_5=a_1*k^4=1$$. Now, if $$a_1<1$$, then $$k>1$$, and all terms from $$a_5$$ ($$a_6$$, $$a_7$$, $$a_8$$, and $$a_9$$), so 4 terms will be greater than 1 AND if $$a_1>1$$, then $$k<1$$, and all terms till $$a_5$$ ($$a_2$$, $$a_3$$ and $$a_4$$), so again 4 terms will be greater than 1. Sufficient.

Hi Bunuel,

Why are we even considering k<1 as the question explicitly says pos constant...?

Also can u please give me an example where k>1 and a>1 gives a^5=1
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Posts: 44647
Re: The next number in a certain sequence is defined by multiply [#permalink]

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22 Apr 2018, 21:31
zanaik89 wrote:
Bunuel 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?

Given that:
$$a_2=a_1*k$$
$$a_3=a_1*k^2$$
$$a_4=a_1*k^3$$
...
$$a_n=a_1*k^{n-1}$$

Also given that k>0 and n=9.

(1) The ninth term in this sequence is 81 --> $$a_9=a_1*k^8=81$$. If $$a_1=1$$, then all but $$a_1$$ will be greater than 1, but if $$a_1=2$$, then all will be greater than 1. Not sufficient.

(2) The fifth term in this sequence is 1 --> $$a_5=a_1*k^4=1$$. Now, if $$a_1<1$$, then $$k>1$$, and all terms from $$a_5$$ ($$a_6$$, $$a_7$$, $$a_8$$, and $$a_9$$), so 4 terms will be greater than 1 AND if $$a_1>1$$, then $$k<1$$, and all terms till $$a_5$$ ($$a_2$$, $$a_3$$ and $$a_4$$), so again 4 terms will be greater than 1. Sufficient.

Hi Bunuel,

Why are we even considering k<1 as the question explicitly says pos constant...?

Also can u please give me an example where k>1 and a>1 gives a^5=1

1. We are told that k is positive constant and k ≠ 1. Nowhere we are told that k is an integer. So, 0 < k < 1 or k > 1.

2. I guess, you are talking about the second statement. $$a_5=a_1*k^4=1$$ cannot be true if both k and a1 are greater than 1. That's why the solution considers two cases: $$a_1<1$$, $$k>1$$ and $$a_1>1$$, $$k<1$$.
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Re: The next number in a certain sequence is defined by multiply   [#permalink] 22 Apr 2018, 21:31

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