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
Source: Manhattan Prep
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A certain series is defined by the following recursive rule:
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BTGmoderatorDC
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Brent@GMATPrepNow
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I'm not crazy about this question.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
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
\(\begin{align}
S_n &= kS_{n-1} \\
S_{25} &= 192 = k S_{24} \\
&= k^{24}*64
\end{align}\)
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__.
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__.
