\(S\) is a sequence \(s_1, s_2, s_3, \ldots, s_n\) in which every term after the first is one less than three times the previous term. If \(s_5 - s_3 = 28,\) which of the following is the first term in the sequence?
A. \(2/3\)
B. \(8/9\)
C. \(1\)
D. \(5/3\)
E. \(2\)
[spoiler]OA=B[/spoiler]
Source: Princeton Review
\(S\) is a sequence \(s_1, s_2, s_3, \ldots, s_n\) in which every term after the first is one less than three times the
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$$Let\ s_1=x$$
$$s_2=3x-1$$
$$s_3=3\left(3x-1\right)-1=9x-3-1=9x-4$$
$$s_4=3\left(9x-4\right)-1=27x-12-1=27x-13$$
$$s_5=3\left(27x-13\right)-1=81x-39-1=81x-40$$
$$Given\ that\ \ s_5-s_3=28$$
$$\left(81x-40\right)-\left(9x-4\right)=28$$
$$81x-40-9x+4=28$$
$$81x-9x=28+36$$
$$\frac{72x}{72}=\frac{64}{72}$$
$$x=\frac{64}{72}=\frac{16}{18}=\frac{8}{9}$$
$$Answer\ =\ B$$
$$s_2=3x-1$$
$$s_3=3\left(3x-1\right)-1=9x-3-1=9x-4$$
$$s_4=3\left(9x-4\right)-1=27x-12-1=27x-13$$
$$s_5=3\left(27x-13\right)-1=81x-39-1=81x-40$$
$$Given\ that\ \ s_5-s_3=28$$
$$\left(81x-40\right)-\left(9x-4\right)=28$$
$$81x-40-9x+4=28$$
$$81x-9x=28+36$$
$$\frac{72x}{72}=\frac{64}{72}$$
$$x=\frac{64}{72}=\frac{16}{18}=\frac{8}{9}$$
$$Answer\ =\ B$$
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Solution:Gmat_mission wrote: ↑Wed Jun 24, 2020 8:08 am\(S\) is a sequence \(s_1, s_2, s_3, \ldots, s_n\) in which every term after the first is one less than three times the previous term. If \(s_5 - s_3 = 28,\) which of the following is the first term in the sequence?
A. \(2/3\)
B. \(8/9\)
C. \(1\)
D. \(5/3\)
E. \(2\)
[spoiler]OA=B[/spoiler]
Let can let s1 = x, so in terms of x, we see that
s2 = 3x - 1,
s3 = 3(3x - 1) - 1 = 9x - 4,
s4 = 3(9x - 4) - 1 = 27x - 13, and
s5 = 3(27x - 13) - 1 = 81x - 40.
Since s5 - s3 = 28, in terms of x, we have:
81x - 40 - (9x - 4) = 28
72x - 36 = 28
72x = 64
x = 64/72 = 8/9
Answer: B
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