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\(S\) 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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\(S\) 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

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$$Let\ s_1=x$$
$$s_2=3x-1$$
$$s_3=3\left(3x-1\right)-1=9x-3-1=9x-4$$
$$s_4=3\left(9x-4\right)-1=27x-12-1=27x-13$$
$$s_5=3\left(27x-13\right)-1=81x-39-1=81x-40$$
$$Given\ that\ \ s_5-s_3=28$$
$$\left(81x-40\right)-\left(9x-4\right)=28$$
$$81x-40-9x+4=28$$
$$81x-9x=28+36$$
$$\frac{72x}{72}=\frac{64}{72}$$
$$x=\frac{64}{72}=\frac{16}{18}=\frac{8}{9}$$
$$Answer\ =\ B$$

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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]

Solution:

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

Scott Woodbury-Stewart
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scott@targettestprep.com

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