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f(n) = [(1)n] × c × n, where c is a positive constant. Difference
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20 Nov 2024, 05:53
20 Nov 2024, 05:53
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40% (01:46) correct
60% (01:57) wrong based on 10 sessions
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\( f(n) = [(–1)^n] × c × n\), where ‘c’ is a positive constant. Difference between the largest and smallest among f(1), f(2) and f(3) is 20.
Quantity A |
Quantity B |
\(f(4)\) |
\(16 \) |
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Re: f(n) = [(1)n] × c × n, where c is a positive constant. Difference
[#permalink]
20 Nov 2024, 06:15
20 Nov 2024, 06:15
3
f(1) = -c
f(2) = 2c
f(3) = -3c
2c + 3c = 20
5c = 20
c = 4.
Therefore f(4) = 1 x 4 x 4 = 16
Answer C
f(2) = 2c
f(3) = -3c
2c + 3c = 20
5c = 20
c = 4.
Therefore f(4) = 1 x 4 x 4 = 16
Answer C
Re: f(n) = [(1)n] × c × n, where c is a positive constant. Difference
[#permalink]
28 Dec 2025, 01:18
28 Dec 2025, 01:18
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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