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P and F are two different sequences, where each upcoming term in seque
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13 May 2024, 03:07
13 May 2024, 03:07
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P and F are two different sequences, where each upcoming term in sequence P is twice the preceding term and each upcoming term in sequence F is three times the preceding term.
First terms for the two sequences are given as \(P_1= 5, F_1= 2\).
First terms for the two sequences are given as \(P_1= 5, F_1= 2\).
Quantity A |
Quantity B |
Max. value of k for \(P_k > F_k\), where \(P_k\) and \(F _k\) are the \(k^{th}\) terms for the two sequences |
\(4\) |
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Re: P and F are two different sequences, where each upcoming term in seque
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13 May 2024, 08:48
13 May 2024, 08:48
1
Sequence P can be written as \(P_k = 2*P_{k-1}\) and Sequence F can be written as \(P_k = 3*P_{k-1}\).
Since \(P_1= 5\), \(P_2 = 2*5=10, P_3 = 2*10=20, P_4 = 2*20=40\)
Since \(F_1= 2\), \(F_2 = 3*2=6, F_3 = 3*6=18, F_4 = 3*18=54\)
Since 40 < 54, \(P_4 < F_4\). Thus, k must be less than 4 for \(P_4 > F_4\) to be true.
Since \(P_1= 5\), \(P_2 = 2*5=10, P_3 = 2*10=20, P_4 = 2*20=40\)
Since \(F_1= 2\), \(F_2 = 3*2=6, F_3 = 3*6=18, F_4 = 3*18=54\)
Since 40 < 54, \(P_4 < F_4\). Thus, k must be less than 4 for \(P_4 > F_4\) to be true.
Re: P and F are two different sequences, where each upcoming term in seque
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19 Apr 2026, 12:39
19 Apr 2026, 12:39
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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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