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# Math Revolution and GMAT Club Contest! There is a sequence An when n

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Intern
Joined: 10 Sep 2012
Posts: 22
GMAT 1: 700 Q49 V35
Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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07 Dec 2015, 06:36
1
Given A 1 =a , A 2 =b , and A n+2 =A n+1 ∗A n

Sequence can be written as - a, b, ab, ab2, a2b3,a3b5... the 6th term i.e A 6 = a3b5
We need to find out whether A 6 <0

Now analyzing statement 1
a < 0
a is negative so a3 is negative, however we don't know the sign of b, so statement 1 is not sufficient.

Analyzing statement 2
ab < 0
either a or b is negative (but not both)
if a is - and b is +, then a3b5 is -
if a is + and b is -, then a3b5 is -
Hence in both cases, a3b5 is negative, which means we can state for sure that A 6 is less than 0. Sufficient.

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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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07 Dec 2015, 08:23
1
A1 = a; A2= b & An+2 = An+1 x An
=> A3 = A2 x A1 = b x a = ab
=> A4 = A3 x A2 = ab x b = a x b^2
=> A5 = A4 x A3 = axb^2 x ab = a^2xb^3
=> A6 = A5 x A4 = a^2xb^3 x axb^2 = a^3xb^5

Case 1 : a<0 => A6 has odd power for for a -ve 'a'. Sufficient
Case 2 : ab<0 => Either a or b is -ve. Both have odd powers hence A6 will always be -ve.

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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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07 Dec 2015, 10:29
1
QUESTION #5:

There is a sequence An when n is a positive integer such that A1=a, A2=b, and An+2=An+1∗An. Is A6<0?
(1) a < 0
(2) ab < 0

Solution:
A6=A5XA4=a^2 X b^3 X ab^2=a^3b^5
A5=A4XA3=ab^2 X ab=a^2 b^3
A4=A3XA2=ab X b=ab^2
A3=A2XA1=ab

A6=a^3b^5

we have to prove whether A6<0 ?
statement 1) a<0: : If a<0, (means: -a) then for the value of +b, value of A6 will be less than zero, but for the value of -b, A6 will be greater than zero . so hereby data is insufficient.
Statement (2) ab < 0: If ab is less than zero then one value between a and b must be negative. if a is negative and b is positive then the value of A6 is less than zero. If a is positive and b is negative, then the value of A6 will be less than zero.So hereby, data is sufficient.

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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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08 Dec 2015, 02:31
1
A1=a, A2=b, and An+2=An+1∗An. Is A6<0?

$$A3 = ab$$
$$A4 = ab^2$$
$$A5 = a^2b^3$$
$$A6 = a^3b^5$$

Since both exponents are odd, we need to know the signs of both a and b to determine A6

1) Insufficient, if b is positive the answer is YES but NO if b is negative
2)Sufficient, only 1 of the two can be negative so the answer is YES

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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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08 Dec 2015, 10:47
1
B. Statement (2) alone is sufficient.

Given,
$$A_1$$=a, $$A_2$$=b
$$A_{n+2}$$ = $$A_n$$($$A_{n+1}$$)

Now,
$$A_3$$ = $$A_{1+2}$$ = $$A_1$$($$A_2$$) = ab
$$A_4$$ = $$A_{2+2}$$ = $$A_2$$($$A_3$$) = (ab) * b = a$$b^{2}$$
$$A_5$$ = $$A_{3+2}$$ = $$A_3$$($$A_4$$) = (a$$b^{2}$$) * ab = $$a^{2}$$$$b^{3}$$
$$A_6$$ = $$A_{4+2}$$ = $$A_4$$($$A_5$$) = ($$a^{2}$$$$b^{3}$$) * (a$$b^{2}$$) = $$a^{3}$$$$b^{5}$$

We get, $$A_6$$ = $$a^{3}$$$$b^{5}$$

(1) a is negative; b could either be positive or negative and therefore we cannot conclude $$a^{3}$$$$b^{5}$$ is negative. Insufficient!
(2) ab is negative; Since we know only one of them is negative, we know that $$a^{3}$$$$b^{5}$$ will always be negative. Sufficient!
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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09 Dec 2015, 05:17
1
Consider A1 and A2 to be +ve => A6 is also positive
If A1 and A2 both are -ve
A3 = -ve * -ve = +ve
A4 = +ve * -ve = -ve
A5 = -ve - +ve = -ve
A6 = -ve * -ve = +ve => A6 is positive

Now to Statement 1
a = A1 < 0 => a is -ve
Since we have evaluated b as -ve, let evaluate b as +ve now
A1 = -ve
A2 = +ve
A3 = +ve * -ve = -ve
A4 = -ve * +ve = +ve
A5 = +ve * -ve = -ve
A6 = -ve * +ve = -ve => A6 is negative

Therefore Not sufficient

Statement 2
ab <0 => a < 0 or b < 0
if a < and b > 0, => A6 is negative (by above)

if a is +ve and b is -ve
A1 = +ve
A2 = -ve
A3 = -ve * +ve = -ve
A4 = -ve * -ve = +ve
A5 = +ve * -ve = -ve
A6 = -ve * +ve = -ve => A6 is negative

Therefore ab < 0 is sufficient

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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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09 Dec 2015, 08:52
1
A3 = A2*A1 = (ab)
A4 = A3*A2 = (ab)b
A5 = A4*A3 = (ab)b(ab) = (b) * (ab)^2
A6 = A5*A4 = (b) * ((ab)^2) * (ab)(b) = (b^2) * ((ab)^3)

Is A6 < 0 can be rewritten as : is (b^2) * ((ab)^3) < 0 ?

Since b^2 is always non negative ; rephrased question will be is ab < 0?

Option B answers the rephrased question directly.

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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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09 Dec 2015, 09:18
1
There is a sequence An when n is a positive integer such that A1=a, A2=b, and An+2=An+1∗An. Is A6<0?

(1) a < 0
(2) ab < 0

A- statement 1 a<0, doesn't tell us type of sequence INSUFFICIENT
B- statement ab<0, atleast one is +ve and other is -ve.
consider terms to solve
a = -1, b= 1; sequence becomes -1, 1, -1, -1, 1, -1
a = 1, b= -1; sequence becomes 1, -1, -1, 1, -1, -1,
a6 is -ve in both of the cases. Hence statement 2 is sufficient
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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10 Dec 2015, 11:57
1
So this is the first time posting on this site, so I apologize in advance if it doesn't look as polished as some of the other posts from more experienced members.

The answer I came up with is B.

The method I took was to work out what A6 is by plugging in what we already know.

A3 = A2*A1
A4 = A3*A2
A5 = A4*A3
A6 = A5*A4

This ends up giving you B^5 * A^3. Since we know that a base to an odd power is going to be the same sign as the base, i.e. positive base remains positive, negative base remains negative, then we know that A6 will be negative if and only if one of these bases, A or B, is negative and the other is positive.

Answer choice A only tells us about one of the bases, so it is wrong. B, on the other hand, does tell us about both. The product of two numbers will be less than zero if, and only if, one of the numbers if positive and the other is negative.

B.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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11 Dec 2015, 03:59
1

A6 is a^3*b^5. Therefore, we need to know the power of both a and b. a<0 does not give us any information regarding the value of b. Hence, not sufficient. For ab<0, A6 can be written as (a^3*b^3)*b^2. Now ab<0 implies a^3*b^3 is <0. and b^2 would always be >0. Hence A6<0 and therefore sufficient.

Let me know if my logic is correct. Cheers
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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13 Dec 2015, 00:32
1
QUESTION #5:

There is a sequence An when n is a positive integer such that A1=a, A2=b, and An+2=An+1∗An. Is A6<0?

(1) a < 0
(2) ab < 0

$$A_1=a, A_2=b$$
--> $$A_3=b*a$$
--> $$A_4=(b*a)*b=b^2*a$$
--> $$A_5=(b^2*a)*(b*a)=b^3*a^2$$
--> $$A_6=(b^3*a^2)*(b^2*a)=b^5*a^3$$ --> The powers of a and b are odd --> $$A_6$$ has the same sign as $$(a*b)$$ does

(1) $$a< 0$$ --> cannot define the sign of $$a*b$$ --> insufficient
(2) $$a*b< 0$$ --> $$A_6 < 0$$ --> sufficient

Asnwer: B
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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06 Dec 2015, 16:25
A1 = a
A2 = b
A3= ab
A4 = ab^2
A5 = a^2b^3
A6 = a^3b^4
Regardless of whether b is positive or negative b^4 will be positive'
A6 will be negative if a is negative
A definitely provides the answer. Sufficient So B, C and E choices are eliminated
ab<0 means one of either a or b are negative
If a is negative then A6 is negative
If b is negative then A6 is positive
This eliminates B
We are left with A
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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08 Dec 2015, 04:29

Statement1:a<0

An+2=An+1*An
A6=A5*A4
A3=a*b
A4=a*b^2
A5=a^2*b^3
A6=a^3*b^4

When a<0, a^3 is -ve, b^4 is always +ve irrespective of b is +ve or -ve , therefore A6 =a^3*b^4 is <0. Statement 1 is sufficient

Statement 2:
A6 can be written as (ab)^3*b.
When ab is <0, then
1)a is -ve and b is +ve --> A6=-ve
2) b is -ve and a is +ve. -->A6=+ve
Not sufficient
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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08 Dec 2015, 10:25
The sequence is: n is positive, so,
A3=A2*A1=ab [when n=1]
A4=A3*A2=ab^2 [when n=2]
A5=A4*A3=a^2*b^3 [when n=3]
A6=A5*A4=a^3*b^4 [when n=4]
Statement 1: it says a is negative. If a is negative, then A6 must be negative since b is always positive. Sufficient.
Statement 2: Either a or b is negative. We don't know which one. Insufficient.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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13 Dec 2015, 02:10

Using the criteria given, A6=a^3*b^5

Condition 1:-
if a<0 then b can either be >0 or <0
if b>0 then A6<0
if b<0 then A6>0
Hence Insufficient

Condition 2:-
ab<0
Alright let's see,
I can write A6 as (ab)^3*b^2
So basically the sign of A6 will be dependent on sign of (ab) because b^2 will always be positive.
Hence Sufficient.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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14 Dec 2015, 08:06
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!

QUESTION #5:

There is a sequence $$A_n$$ when n is a positive integer such that $$A_1=a$$, $$A_2=b$$, and $$A_{n+2}=A_{n+1}*A_n$$. Is $$A_6<0$$?

(1) a < 0
(2) ab < 0

Check conditions below:

Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum

We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth \$299!

All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

MATH REVOLUTION OFFICIAL SOLUTION:

If we alter the original condition and question: $$A_3=A_2×A_1=ab$$, $$A_4=A_3×A_2=(ab)b=a(b^2)$$, $$A_5=A_4×A_3=(ab^2)(ab)=a^2b^3$$ and $$A_6=A_5×A_4=(a^2b^3)(ab^2)=a^3b^5$$. The result is $$A_6=a^3b^5<0$$? → ab<0?. (A square of every number is a positive number. Even if we divide both sides by $$a^2b^4$$ the direction of the inequality does not change.)

So, when we look at 2), ab<0, which means “yes” to question. This is sufficient. Therefore, the correct answer is B.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n  [#permalink]

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24 Dec 2015, 06:25
Sixth term is equal to a^3 * b^5

Statement 1, although suggests that a is negative and odd power of negative is negative, it does not tell us anything about b. hence insufficient

Statment 2 can be interpreted as

ab<0

that implies (a < 0 and b > 0 ) or (a>0 and b<0) i.e in either cases the polarity of a and b are opposite to each other. Since both are odd powers in the sixth term, Yes the sixth term is less than zero.

Hence B
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Re: Math Revolution and GMAT Club Contest! There is a sequence An when n   [#permalink] 24 Dec 2015, 06:25

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