## In the sequence above each term after the first one-half the previous term. If $$x$$ is the tenth term of the sequence,

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### In the sequence above each term after the first one-half the previous term. If $$x$$ is the tenth term of the sequence,

by VJesus12 » Tue Jun 30, 2020 8:08 am
$$\dfrac12, \dfrac14, \dfrac18, \dfrac1{16}, \dfrac1{32}, \ldots$$

In the sequence above each term after the first one-half the previous term. If $$x$$ is the tenth term of the sequence, then $$x$$ satisfies which of the following inequalities?

A) $$0.1 < x < 1$$
B) $$0.01 < x < 0.1$$
C) $$0.001 < x < 0.01$$
D) $$0.0001 < x < 0.001$$
E) $$0.00001 < x < 0.0001$$

[spoiler]OA=D[/spoiler]

Source: GMAT Prep

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### Re: In the sequence above each term after the first one-half the previous term. If $$x$$ is the tenth term of the sequen

by Jay@ManhattanReview » Wed Jul 01, 2020 12:40 am
VJesus12 wrote:
Tue Jun 30, 2020 8:08 am
$$\dfrac12, \dfrac14, \dfrac18, \dfrac1{16}, \dfrac1{32}, \ldots$$

In the sequence above each term after the first one-half the previous term. If $$x$$ is the tenth term of the sequence, then $$x$$ satisfies which of the following inequalities?

A) $$0.1 < x < 1$$
B) $$0.01 < x < 0.1$$
C) $$0.001 < x < 0.01$$
D) $$0.0001 < x < 0.001$$
E) $$0.00001 < x < 0.0001$$

[spoiler]OA=D[/spoiler]

Source: GMAT Prep
Given that the terms are $$\dfrac12, \dfrac14, \dfrac18, \dfrac1{16}, \dfrac1{32}, \ldots$$, we can write them as

$$\dfrac12, \dfrac1{2^2}, \dfrac1{2^3}, \dfrac1{2^4}, \dfrac1{2^5}, \ldots$$

Thus, the 10th term = $$x = \dfrac1{2^{10}} = \dfrac1{1,024} < \left[\dfrac1{1,000} = \dfrac1{10^3} = 0.001 \right]$$

Similarly, the 10th term = $$x = \dfrac1{2^{10}} = \dfrac1{1,024} > \left[\dfrac1{10,000} = \dfrac1{10^4} = 0.0001 \right]$$