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For the zero solution then all the values in the sequence will be zero. If all the values are zero, should we still treat that as solution? Please confirm. Please post OA.
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Aleehsgonji
from 1,
xi = x(i-1)/2
we cannot arrive at solution
from 2
x5= x4/(x4 +1)
but we are unsure whether the same sequence is followed for the other numbers in the series.

combining 1 and 2
x5 = x4/2
x5= x4/(x4 +1)
solving we get x4 = 0 or 1
if x4 = 0 then all the terms will be zero, so x4 =1.
x4 = 1, x3 = 2, x2=4, x1 =8
So answer is "C"

Can you explain how did u get x4 =0 because when we solve these equations we get x4/2= x4/ x4 +1, and now if we simply cancel x4 in numerator then we get x4+1 = 2 , so now My question here is in equations like these shall we assume that answer can be 0 by looking at the initial equation or shall we simplify the equation and then look at the solution. ?

Please advise !

Thanks
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Taking statement 1: It's obvious that we cannot determine X1 from that; however, we can deduce the following:
X2 = X1/2
X3 = X2/2
X4 = X3/2
etc.

Taking statement 2: There is nothing there that says X1 so there is no way we can get X1 from it alone.

Combining both statements: From the sequence in statement 1, we can tell that X5 = X4/2. Now, since we have another definition for X5 in statement 2 [i.e. X5 = X4/(X4+1)], we can equate both defnitions of X5 as follows:

X4/2 = X4/(X4+1)

Since the numerators are the same, we can then conclude from the denominators that X4+1 = 2, and therefore X4 = 1.

Take this information back to the sequence from statement 1, and you can find X3=3, then X2=4, and then X1=8.

Answer is C; you can solve with both statements, but not with any statement independetly.

Hope this helps.
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Given its a sequence of positive numbers, wonder if we should even consider 0 , hence i would exclude x1 as 0.
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C, you need both to derive X1, and 0 is not a positive number...
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skpMatcha
Given its a sequence of positive numbers, wonder if we should even consider 0 , hence i would exclude x1 as 0.

agree,

Sequence is of positive numbers (Xi<>0), Hence C.
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OA is 'C'. Thanks all.



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