In the sequence of positive numbers x1, x2, x3, ..., what is the value

why is it so that we are doing the (1/2) 3 times?

Thanks Bunuel,

But how do you account for the fact that x4 could be equal to zero.

By taking both the statements together, one of the solutions is also x4 = 0. It nowhere mentions in the question that the sequence has all distinct numbers. Or may be I am unaware that sequence is meant to consist of distinct numbers only.

In the sequence of positive numbers X1, X2, X3, ..... What is the value of X1

(1) Xi = Xi-1 / 2 for all integers i > 1

(2) X5 = X4 / X4+1



Guys, I need to know how to solve this question.. Thanks..

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.

last part is not clear. (1/2)^3 how did u get it?

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.

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

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.

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?

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.

How do we know its a GP question?