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A certain series is defined by the following recursive rule:

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by BTGmoderatorDC » Thu Apr 11, 2019 8:41 pm

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A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, wha is the 9th term?

A. ROOT 2
B. ROOT 3
C. 64*ROOT 3
D. 64*3^1/3
E. 64*3^24

OA D

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by Brent@GMATPrepNow » Fri Apr 12, 2019 4:46 am
BTGmoderatorDC wrote:A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, what is the 9th term?

A. ROOT 2
B. ROOT 3
C. 64*ROOT 3
D. 64*3^1/3
E. 64*3^24

OA D

Source: Manhattan Prep
I'm not crazy about this question.
Typically, when we talk about series, Sn represents the SUM of the first n terms of that series.
However, based on the official answer, Sn seems to represent the value of TERM n.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Fri Apr 12, 2019 10:46 am
\(\begin{align}
S_n &= kS_{n-1} \\
S_{25} &= 192 = k S_{24} \\
&= k^{24}*64
\end{align}\)
And
\(S_{1} = 64 = k S_0\)

\(\Rightarrow k^{24} = \frac{192}{64} = 3 \quad \cdots \quad (1)\)

Now, \(S_9 = k^8 * 64\)

From \((1)\), \((k^8)^3 = 3\)

\(k^8 = 3^{\frac{1}{3}}\)

\(S_9 = 3^{\frac{1}{3}} * 64\)

Therefore, answer __D__.
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