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In an arithmetic sequence, each term after the first is the sum of the term before it plus a constant. In the arithmetic

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by VJesus12 » Fri Feb 11, 2022 4:44 am
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In an arithmetic sequence, each term after the first is the sum of the term before it plus a constant. In the arithmetic sequence \(a, b, c, d, e,\) each term after the first is the sum of the preceding term and the constant \(k.\) Which of the following is NOT an arithmetic sequence?

A. \(a, b - 1, c - 2, d - 3, e - 4\)

B. \(a - 1, b - 1, c - 1, d- 1, e - 1\)

C. \(a- 1, b - 2, c - 3, d- 4, e - 5\)

D. \(a, 2b, 3c, 4d, 5e\)

E. \(3a, 3b, 3c, 3d, 3e\)

Answer: D

Source: Princeton Review
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Re: In an arithmetic sequence, each term after the first is the sum of the term before it plus a constant. In the arithm

by swerve » Fri Feb 11, 2022 1:36 pm
VJesus12 wrote: ↑
Fri Feb 11, 2022 4:44 am
 In an arithmetic sequence, each term after the first is the sum of the term before it plus a constant. In the arithmetic sequence \(a, b, c, d, e,\) each term after the first is the sum of the preceding term and the constant \(k.\) Which of the following is NOT an arithmetic sequence?

A. \(a, b - 1, c - 2, d - 3, e - 4\)

B. \(a - 1, b - 1, c - 1, d- 1, e - 1\)

C. \(a- 1, b - 2, c - 3, d- 4, e - 5\)

D. \(a, 2b, 3c, 4d, 5e\)

E. \(3a, 3b, 3c, 3d, 3e\)

Answer: D

Source: Princeton Review
We have arithmetic sequence:
\(a\)
\(b = a+k\)
\(c = a+2k\)
\(d = a+3k\)
\(e = a+4k\)

Checking options.
A.
\(b-1 = a+k-1\)
\(c-2 = a+2k-2\)
\(d-3 = a+3k-3\)
\(e-4 - a+4k-4\)

So the constant here is \(k-1.\) Yes, it is an arithmetic sequence. And so on.

Therefore, D

It will be faster if we just check on numbers. \(a = 1, b = 2, c = 3, d = 4, e = 5.\)
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