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The sequence \(a_1, a_2, \ldots, a_n,\ldots\) is such that \(a_n=\sqrt{a_{n-1}\cdot a_{n-3}}\) for all integers

Expert replies
by Vincen » Thu Jul 30, 2020 8:41 am

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Answers

A

B

C

D

E

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The sequence \(a_1, a_2, \ldots, a_n,\ldots\) is such that \(a_n=\sqrt{a_{n-1}\cdot a_{n-3}}\) for all integers \(n\ge 4.\) If \(a_4=16,\) what is the value of \(a_6?\)

(1) \(a_1=2\)
(2) \(a_2=4\)

[spoiler]OA=C[/spoiler]

Source: Veritas Prep
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Source: — Data Sufficiency |

$$Given\ that\ a_n=\sqrt{\left(a_{n-1}\right)\cdot\left(a_{n-3}\right)}\ \ when\ n\ge4$$ $$a_4=\sqrt{\left(a_{4-1}\right)\cdot\left(a_{4-3}\right)}\ \ =\sqrt{a_3\cdot a_1}=16$$
Target question=> What is the value of
$$a_6=\sqrt{\left(a_{6-1}\right)\cdot\left(a_{6-3}\right)}\ \ =\sqrt{a_5\cdot a_3}$$
$$Where\ \ a_5=\sqrt{a_4\cdot a_2}=\sqrt{16\cdot a_2}$$
$$Statement\ 1:\ a_1=2$$
$$From\ the\ question\ stem,\ a_4=\sqrt{a_3\cdot a_1}$$
$$16^2=\left(\sqrt{a_3\cdot a_1}\right)^2$$
$$16^2=a_3\cdot2$$
$$a_3=\frac{256}{2}=128$$
$$a_6=\sqrt{a_5\cdot a_3}\ and\ a_5=\sqrt{a_4\cdot a_2}=\sqrt{16\cdot a_2}$$
$$Here,\ a_2\ is\ unknown.\ $$
So, therefore, the target question cannot be answered. Hence, statement is NOT SUFFICIENT.
$$Statement\ 2:\ a_2=4$$
$$a_6=\sqrt{a_5\cdot a_3}\ and\ a_5=\sqrt{a_4\cdot a_2}=\sqrt{16\cdot 4}$$ $$a_5=\sqrt{16\cdot4}=\sqrt{64}=8$$
$$a_6=\sqrt{8\cdot a_3}\ but\ a_{3\ }is\ unknown.$$
Therefore, the target question cannot be answered. Hence, statement 2 is also NOT SUFFICIENT.

Combining both statements together;
$$From\ statement\ 1=>a_1=2,\ a_3=128$$
$$From\ statement\ 2=>a_2=4,\ a_5=8$$
$$From\ question\ stem=>a_4=16$$
$$Therefore,\ a_6=\sqrt{a_5\cdot a_3}=\sqrt{8\cdot128}$$
$$a_6=\sqrt{1024}=32$$
However, combining both statements together are SUFFICIENT. Hence, validating option C as the correct answer.
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