In the sequence of positive numbers \(x_1\), \(x_2\), \(x_3\), ..., what is the value of \(x_1\)?

(1) \(x_i=\frac{x_{(i-1)}}{2}\) for all integers \(i>1\) --> we have the general formula connecting two consecutive terms (basically we have geometric progression with common ratio 1/2), but without the value of any term this info is insufficient to find \(x_1\).

(2) \(x_5=\frac{x_4}{x_4+1}\) --> we have the relationship between \(x_5\) and \(x_4\), also insufficient to find \(x_1\) (we cannot extrapolate the relationship between \(x_5\) and \(x_4\) to all consecutive terms in the sequence).

(1)+(2) From (1) \(x_5=\frac{x_4}{2}\) --> \(\frac{x_4}{2}=\frac{x_4}{x_4+1}\) --> \(x_4=1\) --> \(x_4=1=x_1*(\frac{1}{2})^3\) --> \(x_1=8\). Sufficient.

Answer: C.
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\(x_i=\frac{x_{(i-1)}}{2}\), so every next term is preivious term times \(\frac{1}{2}\) --> \(x_4=x_3*\frac{1}{2}=x_2*\frac{1}{2}*\frac{1}{2}=x_1*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=x_1*(\frac{1}{2})^3\).
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Stem says: "In the sequence of positive numbers ..."
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from 1, simply put the values of i = 2,3 or 4 but we can not find the value of X1

we can only know

X2 = X1 / 2 or
X3 = X2 / 2 = X1 / 4 or
X4 = X1 / 8 or
X5 = X1 / 16 etc....
insufficient

from 2, X5 = X4 / X4+1
cannot find X1 , insufficient

combine 1+2, X5 = X1 / 16
and X4 = X1 / 8

so, X1 / 16 = (X1 / 8)/((X1 / 8)+1)

can find X1 hence sufficient

Note : you will find two values of X1 from above quadratic equation i.e X1 = 0 and X1 = 8,
since it is given that X1 is positive so we cant take X1=0 hence sufficient

Please read the red part carefully.

Or from \(x_i=\frac{x_{(i-1)}}{2}\):

\(x_2=\frac{x_1}{2}\);

\(x_3=\frac{x_2}{2}=\frac{x_1}{4}\);

\(x_4=\frac{x_3}{2}=\frac{x_1}{8}\).

Hope it's clear.
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I tried to do this by plugging in numbers and after 3+ minutes, I chose "Neither Suff" b/c I wasn't seeing any trends.

A and B are both insufficient because we don't have a starting point. When I combined, I still didn't see a starting point and went through the calculations only to realize that I STILL didn't have a starting point.

In retrospect, if I was to plug in numbers, what would be the better approach? Should I pick a number for say X5 and work backwards on both A and B or does picking numbers in this problem fail? It seemed to me that I could get MULTIPLE values since I could assign ANY value to X5 etc.?

Bunuel: to get x4 = 1 do we cross multiply? Can you show the steps to attain this value ?

This is not my strongest topic, but I just want to share my experience with this type of questions.
One can solve this type of questions if you have a general formula (sometimes given in the question stem) and concrete values for at least two other terms in the sequence.

Question: We are not given any formula here, we are just asked to find a value of X1.
(1) It's just a general formula, that gives us the relaionship of any two terms in the sequence, but we don't have any concrete values. Not sufficient
(2) Here we are given ONLY a relationship between X5 and X4, but you cannot just use this relationship for other terms, X4 and X5 could by any positive values. Not sufficient.
(1) + (2): x4=x4/x4+1 = x4/2, ok, so you can find the x4, in the same manner you can find x3 etc. so , STOP, there is no need to calculate further, one can see here, that it's possible to find x1
Answer C

Would appreciate some comments from math experts. I have seen many questions of this type, and there are always about a general formula and some concrete values, which we can use to find any term in the sequence.
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Bunuel I got this wrong because I thought x1 = 0 is also a solution. But since the question says it is a sequence of positive numbers, I guess I cannot assume that.

On a slightly different note, can the values in a sequence be a constant?

A video with a more thorough explanation can be found here:
https://www.youtube.com/watch?v=3l0g2OPhI_o

Statement (1) gives us a general equation that can be applied to any value of x. Helpful, but not sufficient.

Statement (2) gives us an equation that is specific to x5; it does not necessarily apply to every other value of x. Perhaps helpful, but not sufficient.

In combination, however, we can now write two equations with two unknowns (x5 and x4). Any time you have two unknows, you can solve for both of them if you have two equations that are different from each other (as we do here), so without doing any math we know that we can find a value for x4, and then we could go back and use the equation in statement (1) to find x1.

Answer C.
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