In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created by the equation \(a_n=a_{(n−1)}∗2\). By what percent is the 11th term greater than the 9th term in this sequence?

A. 100%
B. 200%
C. 300%
D. 400%
E. 800%
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Its a Gp series so 11 term = a^10 ; 9th term = a^8
so difference = 3*2^8 => percent gap = 2^8*3/2^8 *100 = 300 %
hence c is correct
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\(a_1 = 1, a_2 = 2, a_3 = 4\)

Using \(a_n=a_{(n−1)}∗2\)

\(a_1 = 1\)
\(a_2 = 2\)
\(a_3 = 4 = (2 * 2)\)
\(a_4 = 8 = (2 * 2 * 2)\)
\(a_5 =16 = (2 * 2 * 2 * 2)\)
\(a_6=32=(2 * 2 * 2 * 2*2)\)

By definition, a geometric sequence has a common ratio.

The geometric expansion from \(a_2\) to \(a_4\) has the same ratio as the geometric expansion from \(a_{212}\) to \(a_{214}\), or the expansion between any two terms, 9\(^{th}\) and 11\(^{th}\) included.

In other words, \(a_{n+2}\) will be the same percent greater than \(a_{n}\) no matter which terms you choose.

So: \(a_4\) is what percent greater than \(a_2\)?

Percent increase = \(\frac{New-Old}{Old} * 100\)

\(\frac{8 - 2}{2} =\frac{6}{2} * 100\) = 300 percent

Answer C
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I would just say that every GM sequence within a GM sequence holds the same percentage difference - so difference between first and third term is same as the difference between 9th and 11th term
--> therefore 300%