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

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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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.

Answer: (B)

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

Answer B
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Answer is A

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

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.
Answer is A.

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!

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

Answer is B

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.

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

Answer - B ,

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

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.

IMO Answer is B

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

A1=a, A2=b, and An+2=An+1∗An.
Is A6<0?

From equation, An+2=An+1∗An
A6=A5*A4
A5=A4*A3
A4=A3*A2
A3=A2*A1

Therefore A6= (A2)^5 * (A1)^3
Now lets take the choices

(1) a < 0
a<0 and b>0, Then A6<0 Yes
But if a<0 ,b<0 Then A6 >0 No

Choice 1 is insufficient.


(2) ab < 0
Means that a and b are of opposite signs . Therefore A6 which is equal to (A2)^5 * (A1)^3 will always be negative and A6<0.

Therefore B is sufficient.

Hence choice B is answer.

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

Answer B

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.

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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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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