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

Following information is given:
1. An is a sequence and n is a positive integer.
2. A1 = a and A2 = b (for n = 1 and 2 values of the sequence are given)
3. An+2 = An+1∗An (this formula or say pattern is given to find the subsequent values)

This pattern tells us that any two consecutive values of the sequence when multiplied give the third consecutive value of the sequence. Hence, Subsequent values of this sequence would be:

when n = 3, A3 = A2 * A1 = ba
when n = 4, A4 = A3 * A2 = ba*b = b2a
when n = 5, A5 = A4 * A3 = b2a*ba = b3a2
when n = 6, A6 = A5 * A4 = b3a2*b2a = b5a3

we are asked if A6 < 0 i.e is A6 negative? i.e. is b5a3 negative?
as we see powers of both b and a are odd, b5a3 will be negative if a and b have opposite signs.

statement-1: a < 0
as b can be positive or negative, b5a3 can be positive or negative.
As we dont have the definite answer with statement-1, its INSUFFICIENT.

statement-2: ab < 0
this tells us that a and b have opposite signs. b5a3 will be negative.
We have definite answer with statement-2. its SUFFICIENT.

Option (B) is correct answer.

A1 = a; A2 = b;
A3 = A2 * A1 = a*b
A4 = A3 * A2 = ab * b = a * b^2
A5 = A4 * A3 = (a * b^2) * (ab) = (a^2 * b^3)
A6 = A5 * A4 = (a^2 * b^3) * (a * b^2) = (a^3 * b^5) = (ab)^3 * b^2

St1: a < 0 --> Not sufficient as 'b' is not known

St2: ab < 0
If ab < 0 then a and b is not equal to 0.
A6 depends upon the sign of (ab) since b^2 is always positive.
So ab < 0 --> A6 is negative
Statement 2 is sufficient.

Answer: B

I will go with B as my choice of answer.

A1= a, A2=b and An+2= An+1 + An. Question asks us is A6 < 0? this can be re framed as Is a^3 X b^5 < 0?

Case 1: a < 0 this implies a is negative. So a^3 is also negative. This depends on sign of b. If b is positive we will get A6 < 0 but if b is negative then b^5 is also negative so whole term A6 will be positive and that means that A6 may be positive or negative depending on the value of b................... Insufficient

Case 2: ab < 0 this implies either a or b is negative. if a is negative, b will be positive so this would imply A6 < 0 and also if b is negative then a will be positive. So in both case the sign will be preserved and will definitely give us sufficient conditions or answer of question A6 < 0...................... Sufficient

Answer is B

From the formula:
\(A_6 = A_5 * A_4 = (A_4)^2 * A_3 = (A_3)^3 * (A_2)^2 = (A_2)^5 * (A_1)^3 = b^5*a^3 = (ab)^3 * b^2\)

1. a < 0 is insufficient because \(A_6\) > 0 when b < 0 & \(A_6\) < 0 when b > 0.
2. ab < 0 is sufficient to conclude that \(A_6<0\) because\((ab)^3<0\) and \(b^2>0\) (b cannot equal 0 since ab<0)

Hence B.

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


On explanding,

A6 = A1^3 * A2*5 = a^3 *b*5

We are asked if a^3 *b*5 < 0
1) a < 0. Insufficent as we have to know whether b is positive or negative.
2) ab< 0. Sufficient since if ab < 0 then multiplication of add powers of a and odd powers of b will also be less than 0.

Ans. B

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


Soln: \(A1 = a, A2 = b,\) and since \(An+2=An+1∗An,\) we can deduce \(A3 = A2*A1 = ab.\)
Similarly \(A4 = A3*A2 = ab^2, A5 = a^2b^3, and A6 = a^3b^5.\)
So, the problem translates to: whether \(a^3b^5 < 0\) ??

1. \(a < 0,\) so, \(a^3 < 0.\) But we do not know b. If b = 1, YES. If b = -1, NO. INSUFFICIENT.

2. \(ab < 0.\) So, \(a & b\) are of opposite sign. If \(a<0\) and \(b>0,\) then \(a^3<0\) and \(b^5>0,\) and so \(A6 < 0.\) YES.
If \(a>0\) and \(b<0,\) then \(a^3>0\) and \(b^5<0,\) and so \(A6 < 0.\) YES. Both YES. SUFFICIENT.

Answer: B.

A1=a,A2=b

An+2=An+1*An

suppose n=1, then A3=A2*A1=ab

A4=A3*A2=ab*b=ab^2

A5=A4*A3=ab^2*ab=a^2b^3

A6=A5*A4=a^2b^3*ab^2=a^3b^5<0.

from above both a and b have odd powers .

Statement 1 if a<0, we cant conclude we don't know value of b

Statement 2
ab<0 means
I a>0 and b<0 or
a<0 and b>0. both value satisfy above A6<0.
so option B is correct

given,
\(A_1\) = a
\(A_2\) = b

from the formula, \(A_{n+2}\) = \(A_{n+1}\)* A,
we get the remaining terms.
\(A_3\) = \(A_2\)* \(A_1\) = a*b = ab
\(A_4\) = \(A_3\)* \(A_2\) = ab * b = a\(b^2\)
\(A_5\) = \(A_4\)* \(A_3\) = a\(b^2\)* ab = \(a^2\)\(b^3\)
\(A_6\) = \(A_5\)* \(A_4\) = \(a^2\)\(b^3\)* a\(b^2\) = \(a^3\)\(b^5\)

Now, we need to know if \(a^3\)\(b^5\) is negative or not
since, the powers of a and b are odd, we have 3 cases. The product can be
1. positive ( if both a & b are of same sign.)
2. negative ( if one of them is negative and other positive)
3. zero. (if any one or both of them are zero)

Stmt 1: a <0
It doesn't give any information about b.
-- b can be positive --- in that case, the product will be negative.
-- b can be negative--- in that case, the product will be positive.
-- b can be zero --- in that case, the product will be zero.
Hence we cant conclude using the statement 1 alone.

stmt 2: ab <0
This is possible only if one of them is negative and the other is positive. Also, it is to be noted that none of them is zero.
Hence we can conclude that the product \(a^3\)\(b^5\) will be negative.

Answer: B

We have a series where every term from the third term onwards is determined by multiplying the previous two terms. If either or both of the previous two terms are 0 then that term (and every following term) will be 0. If the previous two terms are both the same sign then the term we are calculating will be +ve, if the two previous terms are different in sign then the term we are calculating will be -ve.

Since we're looking to determine whether A6<0 based on information about whether a<0 or the product ab<0 let's quickly jot out all of the possible sequences:

ab....A6
++++++
+--+--
--+--+
-+--+-
000000
?00000
0?0000

Condition 1) tells us that a is -ve, which means that the highlighted sequences are possible. Since in the highlighted sequences term A6 can be +ve, -ve or 0, 1) on it's own is not sufficient, we can rule out answers A and D.

Condition 2) tells us that the product ab<0. This means that neither a nor b is 0, and that a and b are of different signs. Therefore the bold sequences are possible. Both of the bold sequences have a -ve A6, therefore 2) is sufficient to answer the question "is A6<0", and the correct answer is B.

A1=a, A2=b, and An+2=An+1∗An.

A6 = A(4+2) = A(4+1).A4

A3= A(1+2) =A2.A1 = ab

A4= A(2+2) = A3.A2 = ab^2

A5= A(3+2) = A4.A3 = a^2b^3

A6= A(4+2) = A5.A4 = a^3b^5 = (ab)^3 b^2..

1) a<0, b+? NS
2) ab<0 .. A6 <0 S

B

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

\(A_3 = A_2*A_1 = b*a\)
\(A_4 = A_3*A_2 = (b*a)*b\)
\(A_5 = A_4*A_3 = ((b*a)*b)*(b*a)\)
\(A_6 = A_5*A_4 = ((b*a)*b)*(b*a)*(b*a)*b = (b*a)^3*b^2\)

Statement (1): a < 0
Since we don't have any info on b statement (1) is insufficient

Statement (2): ab < 0
\(A_6 = (b*a)^3*b^2\) is < 0
Sufficient

Answer (B)

A1= a
A2 = b
A3= ab
A4= ab^2
A5=a^2b^3
A6 = a^3b^5

from statement 1: a<0
we do not know about b is b <0 then a6<0
if b>0 the a6 is not less than 0

in sufficient

statement 2:
ab<0

A6= a^3 B^5
this can be written as
a^2B^4 * ab
a^2B^4 will always be positive
since ab< 0 A6 < 0

Sufficient
Ans: B

A3 = A1*A2 , A4 = A3*A2 => A4 = A1*A2^2 , A5 = A4*A3 => A5 = A1^2 * A2^3 ||ly A6 = A1^3 * A2^5
We have to check if (a^3 * b^5) < 0
1) a<0 but we have no info about b, thus insufficient

2) ab<0 which means either a<0 or b<0 thus A6<0 thus (a^3 * b^5) < 0 . Therefore sufficient

Answer is B

A6=A5*A4
=A4*A3*A3*A2
=A3*A2*A2*A1*A2*A1*A2
=A2*A1*A2*A2*A1*A2*A1*A2
=b*a*b*b*a*b*a*b
=\(b^5*a^3\)

Since both the powers are odd, we only need to find out whether the a*b with odd powers is positive or negative.

1. a<0
Nothing for b
Not Sufficient

2. ab<0
ab with odd powers is given here.
Sufficient

Answer: B

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

A6 = a^3 b^5

(1) Not sufficient as dont know anything about the sign of b
(2) if ab<0 ; then either of the two is negative. That is, both has opposite sign and that is what we need since we have odd powers of a and b in A6. Hence sufficient.

B.

OA: B

Solution:
A1 = a
A2 = b
A3 = a * b
A4 = a * b^2
A5 = a^2 * b^3
A6 = a^3 * b ^ 5

For A6 to be < 0, either a < 0 (and b > 0) or vice versa

(1) a < 0 No info about b, so insufficient.
(2) ab < 0 this is exactly the condition we need as stated above, hence sufficient.