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# In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value

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Joined: 02 Sep 2009
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In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value  [#permalink]

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09 Oct 2018, 03:08
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Question Stats:

65% (01:21) correct 35% (01:46) wrong based on 75 sessions

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In the sequence of non-zero numbers $$a_1$$, $$a_2$$, $$a_3$$, ... $$a_n$$, ..., the value of $$a_{(n+1)} = \frac{a_n}{2}$$, for all positive integers n. What is the value of $$a_6$$?

(1) $$a_2 =\frac{1}{2}$$

(2) $$a_2 − a_6 = \frac{15}{32}$$

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In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value  [#permalink]

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09 Oct 2018, 03:55
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Bunuel wrote:
In the sequence of non-zero numbers $$a_1$$, $$a_2$$, $$a_3$$, ... $$a_n$$, ..., the value of $$a_{(n+1)} = \frac{a_n}{2}$$, for all positive integers n. What is the value of $$a_6$$?

(1) $$a_2 =\frac{1}{2}$$

(2) $$a_2 − a_6 = \frac{15}{32}$$

Question: What is the value of $$a_6$$?

Given: $$a_{(n+1)} = \frac{a_n}{2}$$
i.e. every Next term of the sequence is Half the previous term

ie. in order to answer this question we need any one term of the series and we can find all other terms

Question REPHRASED: What is the Numerical Value of any term of the series?

Statement 1: $$a_2 =\frac{1}{2}$$

SUFFICIENT

Statement 2: $$a_2 − a_6 = \frac{15}{32}$$

Using the series $$a_2$$ may be translated into $$a_6$$ and then we can find the numerical value using this statement hence

$$a_3 = a_2 / 2$$
$$a_4 = a_3 / 2 = a_2/4$$
$$a_5 = a_4 / 2 = a_2/8$$
$$a_6 = a_5 / 2 = a_2/16$$
i.e. $$a_2 = 16*a_6$$

Now, $$a_2 − a_6 = \frac{15}{32}$$
i.e. $$16*a_6 − a_6 = \frac{15}{32}$$
i.e. $$15*a_6 = \frac{15}{32}$$
i.e. $$a_6 = \frac{1}{32}$$

SUFFICIENT

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In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value  [#permalink]

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09 Oct 2018, 07:17
Bunuel wrote:
In the sequence of non-zero numbers $$a_1$$, $$a_2$$, $$a_3$$, ... $$a_n$$, ..., the value of $$a_{(n+1)} = \frac{a_n}{2}$$, for all positive integers n. What is the value of $$a_6$$?

(1) $$a_2 =\frac{1}{2}$$

(2) $$a_2 − a_6 = \frac{15}{32}$$

1) $$a_2 =\frac{1}{2}$$

$$a_3 = \frac{1}{2/2}$$ = $$\frac{1}{4}$$

$$a_6$$will then be in the form of$$\frac{1}{2^n+1}$$

thus $$a_6 = \frac{1}{32}$$

sufficient.

for 2)
$$a_2 = \frac{a_1}{2}$$

$$a_3 = \frac{a_2}{2}$$

etc...

$$a_6 = \frac{a_1}{2^n+1}$$

$$\frac{a_1}{2} - \frac{a_1}{32} = \frac{16*a_1}{32} - \frac{a_1}{32} = \frac{15*a_1}{32}$$

$$\frac{15*a_1}{32} = \frac{15}{32}$$

$$a_1 = 1$$

$$a_6 = \frac{a_1}{2^n+1} = \frac{1}{32}$$

Sufficient

In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value   [#permalink] 09 Oct 2018, 07:17
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