How many of the first nine terms are greater than 1?

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Hello,

Can you please assist with the following question from MGMAT:


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.


OA: B


Thanks a lot for your help - Sri

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by theCodeToGMAT » Sat Oct 12, 2013 10:08 pm
x1 = A
x2 = kA
x3 = K*k*A
..
x9 = k^8 * A

To find
: Terms greater than 1 in first 9 terms

Statement 1:
81 = k^8 * A
INSUFFICIENT

Statement 2:
1 = k^4 * A
From this we can deduce that either K or A is in fraction..
So, either 1,2,3 & 4th Terms or 6,7,8 and 9th terms are greater or lower than 1..
So total "4"
SUFFICIENT

Answer [spoiler]{B}[/spoiler]
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by [email protected] » Sun Oct 13, 2013 2:51 pm
Hi gmattesttaker2,

This DS question tells us that each term in a sequence is equal to the PREVIOUS term multiplied by a POSITIVE CONSTANT.

For example, the sequence 1, 2, 4, 8, 16 would fit this definition (another example would be 16, 8, 4, 2, 1, 1/2, etc.). We don't know any of the terms though and we don't know the constant (it could be either an integer, fraction or mixed number). We DO know that since we're multiplying by a positive constant that the sequence of numbers either "increases" or "decreases."

We're asked how many of the first 9 terms are greater than 1? It's interesting that the question asks how many are greater than 1.

Fact 1: the 9th term = 81

If the terms are 9, 81, 729, etc., then ALL 9 terms are greater than 1.
If the terms are 1, 81, 81(81), etc. then 8 of the germs are greater than 1.
Fact 1 is INSUFFICIENT

Fact 2: the 5th term = 1

Since the sequence either increases or decreases, we'd have...
4 numbers less than 1, 1, 4 numbers greater than 1
or
4 numbers greater than 1, 1, 4 numbers less than 1

Regardless of which option, we end up with exactly 4 terms.
Fact 2 is SUFFICIENT

Final Answer: B

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by gmattesttaker2 » Thu Oct 17, 2013 7:16 pm
[email protected] wrote:Hi gmattesttaker2,

This DS question tells us that each term in a sequence is equal to the PREVIOUS term multiplied by a POSITIVE CONSTANT.

For example, the sequence 1, 2, 4, 8, 16 would fit this definition (another example would be 16, 8, 4, 2, 1, 1/2, etc.). We don't know any of the terms though and we don't know the constant (it could be either an integer, fraction or mixed number). We DO know that since we're multiplying by a positive constant that the sequence of numbers either "increases" or "decreases."

We're asked how many of the first 9 terms are greater than 1? It's interesting that the question asks how many are greater than 1.

Fact 1: the 9th term = 81

If the terms are 9, 81, 729, etc., then ALL 9 terms are greater than 1.
If the terms are 1, 81, 81(81), etc. then 8 of the germs are greater than 1.
Fact 1 is INSUFFICIENT

Fact 2: the 5th term = 1

Since the sequence either increases or decreases, we'd have...
4 numbers less than 1, 1, 4 numbers greater than 1
or
4 numbers greater than 1, 1, 4 numbers less than 1

Regardless of which option, we end up with exactly 4 terms.
Fact 2 is SUFFICIENT

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

Thanks a lot for the explanation. I had a question about Stmnt. 1

(1) The ninth term in this sequence is 81.

Does this mean that the 9th term (say S9) S9 = 81? The part where you have mentioned "We DO know that since we're multiplying by a positive constant that the sequence of numbers either "increases" or "decreases."" is clear. However, I am still not not clear with how 1 is in-sufficient and 2 is sufficient.

Thanks a lot for your help.

Best Regards,
Sri

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by [email protected] » Thu Oct 17, 2013 11:11 pm
Hi Sri,

For Fact 1, since we don't know what the constant "k" is (it could be a positive fraction or a positive number > 1). Here are two examples that prove that the answer to the question changes:

If k = 3, then the terms (working backwards from the 9th term....) are:
81, 27, 9, 3, 1, 1/3, 1/9, 1/27, 1/81
Here, the number of terms greater than 1 = 4

If k = 1/3, then the terms (working backwards from the 9th terms....) are:
81, 243, 729, then they get bigger and bigger.....
Here, the number of terms greater than 1 = 9

Since we have 2 different answers, Fact 1 is INSUFFICIENT.

The big DEDUCTION from this set of TEST cases is that the sequence of numbers MUST either increase (if the constant is > 1) or decrease (if the constant is < 1).

Fact 2: The fifth term is 1.

Since we're asked about the first nine terms, we now know that the middle term is 1.

There are 4 terms BEFORE the 1 and 4 terms AFTER the 1.

If the sequence goes UP, then the numbers AFTER the 1 are greater than 1 and the others are not.
If the sequence goes DOWN, then the numbers BEFORE the 1 are greater than 1 and the others are not.

Either way, there are 4 terms that are greater than 1.

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by gmattesttaker2 » Fri Oct 18, 2013 8:48 pm
[email protected] wrote:Hi Sri,

For Fact 1, since we don't know what the constant "k" is (it could be a positive fraction or a positive number > 1). Here are two examples that prove that the answer to the question changes:

If k = 3, then the terms (working backwards from the 9th term....) are:
81, 27, 9, 3, 1, 1/3, 1/9, 1/27, 1/81
Here, the number of terms greater than 1 = 4

If k = 1/3, then the terms (working backwards from the 9th terms....) are:
81, 243, 729, then they get bigger and bigger.....
Here, the number of terms greater than 1 = 9

Since we have 2 different answers, Fact 1 is INSUFFICIENT.

The big DEDUCTION from this set of TEST cases is that the sequence of numbers MUST either increase (if the constant is > 1) or decrease (if the constant is < 1).

Fact 2: The fifth term is 1.

Since we're asked about the first nine terms, we now know that the middle term is 1.

There are 4 terms BEFORE the 1 and 4 terms AFTER the 1.

If the sequence goes UP, then the numbers AFTER the 1 are greater than 1 and the others are not.
If the sequence goes DOWN, then the numbers BEFORE the 1 are greater than 1 and the others are not.

Either way, there are 4 terms that are greater than 1.

GMAT assassins aren't born, they're made,
Rich

Hello Rich,

Thanks for the excellent explanation. It is clear now. Many thanks again for your valuable time and help.

Best Regards,
Sri

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by eektvas » Mon Jul 04, 2016 6:42 am
what if a =1. then any k multiplied by it will always give 1 / Should mpt c be the answer

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by [email protected] » Mon Jul 04, 2016 9:33 am
Hi eektvas,

It's not clear what you mean when you say a = 1 (since this question does not have any variable called "a"). That having been said, we're told that "k" CANNOT be 1, so whatever that constant is, the sequence of numbers WILL change from term to term.

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by 800_or_bust » Tue Jul 05, 2016 5:49 am
gmattesttaker2 wrote:Hello,

Can you please assist with the following question from MGMAT:


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.


OA: B


Thanks a lot for your help - Sri
(1) Not sufficient. If k=81, then only the ninth term is greater than 1. If k=9, then the 8th and 9th terms are greater than 1. Etc...

(2) Sufficient. Since the 5th term is 1, either the first 4 terms must be greater than 1 (if 0<k<1) or the last 4 terms must be greater than 1 (if k>1). Either way, we have 4 terms that are greater than 1.

Answer: B
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by Matt@VeritasPrep » Fri Jul 08, 2016 2:21 pm
It might be easier to just write this out.

Suppose the first term is x. Then the sequence is

x, x*k, x*k², x*k³, ..., x*k�

From S2, we have x * k� = 1, or x = 1/k�.

Replacing x with 1/k� in the sequence, we find that our sequence is

1/k�, 1/k³, ..., 1, ..., k³, k�

If k > 1, then the last four terms are > 1.

If 1 > k > 0, then the first four terms are > 1.

In either case, we have exactly four terms that are greater than 1.