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The sequence \(a_1, a_2, …, a_n, …\) is such that

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The sequence \(a_1, a_2, …, a_n, …\) is such that

by Gmat_mission » Mon Jun 10, 2019 5:35 am

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

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

[spoiler]OA=C[/spoiler]

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Source: — Data Sufficiency |
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by deloitte247 » Fri Jun 14, 2019 10:55 pm

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$$a_n=\sqrt{a_{n-1}\cdot a_{n-3}}$$
$$n\ge4$$
$$a_6=\sqrt{a_{6-1}\cdot a_{6-3}}=\sqrt{a_5\cdot a_3}$$
$$a_5=\sqrt{a_4\cdot a_2}$$
$$a_3=a_{2+1}=a_2+a_1$$
$$a_4=16$$
$$what\ is\ the\ value\ of\ a_6$$
Statement 1
$$a_1=2$$
$$value\ of\ a_5,a_3\ and\ ,a_2\ are\ not\ known$$
$$Hence,\ we\ can't\ \inf er\ the\ value\ of\ a_6$$
Statement 1 is INSUFFICIENT.

Statement 2
$$a_2=4$$
$$value\ of\ a_5,\ a_3\ and\ a_1\ are\ unknown$$
Statement 2 is INSUFFICIENT.

Combining statement 1 and 2 together
$$a_3=a_2+a_1=4+2=6$$
$$a_5=\sqrt{a_4\cdot a_2}=\sqrt{16\cdot4}=\sqrt{64}=8$$
$$a_5=\sqrt{a_4\cdot a_2}=\sqrt{16\cdot4}=\sqrt{64}=8$$
Both statements combined together are SUFFICIENT.
$$a_6=\sqrt{a_5\cdot a_3}=\sqrt{8\cdot6}=\sqrt{48}=\sqrt{16\cdot3}=4\sqrt{2}$$

$$answer\ is\ Option\ C$$
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