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07 Dec 2015, 07:36
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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.
Answer - B
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.
Answer - B
07 Dec 2015, 09:23
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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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=> 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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07 Dec 2015, 11:29
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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)
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)
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