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P, F , and G are three different geometric sequences, where each
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06 Sep 2026, 14:19
06 Sep 2026, 14:19
Expert Reply
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Question Stats:
50% (01:32) correct
50% (02:01) wrong based on 6 sessions
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P, F , and G are three different geometric sequences, where each upcoming term in sequence P is twice the preceding term, each upcoming term in sequence F is three times the preceding term, and each upcoming term in sequence G is four times the preceding term. The first terms are given as \(P_1=20, F_1=3\) , and \(G_1=1\) .
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Max. value of k for which \(P_k>F_k\) and \(F_k>G_k\) |
3 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
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P, F , and G are three different geometric sequences, where each
[#permalink]
11 Sep 2026, 10:32
11 Sep 2026, 10:32
1
Expert Reply
Solution
We have:
\(P_k = 20(2)^{k-1}\)
\(F_k = 3(3)^{k-1}\)
\(G_k = 1(4)^{k-1}\)
We need the largest k such that both:
\(P_k>F_k\) and \(F_k>G_k.\)
Solve \(P_k>F_k\)
\(20\cdot2^{k-1}>3\cdot3^{k-1}\)
Check values:
\((k=4:\quad P_4=160,\ F_4=81)\) ✓
\((k=5:\quad P_5=320,\ F_5=243)\) ✓
\((k=6:\quad P_6=640,\ F_6=729)\) ✗
So \((k\le5\)\).
2. Solve \(F_k>G_k\)
\(3\cdot3^{k-1}>4^{k-1}\)
Check:
\(k=2:\quad F_2=9,\ G_2=4\) ✓
\(k=3:\quad F_3=27,\ G_3=16\) ✓
\(k=4:\quad F_4=81,\ G_4=64\) ✓
\(k=5:\quad F_5=243,\ G_5=256\) ✗
So \(k\le4\)
Therefore the maximum \(k\) satisfying both conditions is 4.
Compare
Quantity \(A = 4\)
Quantity \(B = 3\)
Quantity A is greater.
We have:
\(P_k = 20(2)^{k-1}\)
\(F_k = 3(3)^{k-1}\)
\(G_k = 1(4)^{k-1}\)
We need the largest k such that both:
\(P_k>F_k\) and \(F_k>G_k.\)
Solve \(P_k>F_k\)
\(20\cdot2^{k-1}>3\cdot3^{k-1}\)
Check values:
\((k=4:\quad P_4=160,\ F_4=81)\) ✓
\((k=5:\quad P_5=320,\ F_5=243)\) ✓
\((k=6:\quad P_6=640,\ F_6=729)\) ✗
So \((k\le5\)\).
2. Solve \(F_k>G_k\)
\(3\cdot3^{k-1}>4^{k-1}\)
Check:
\(k=2:\quad F_2=9,\ G_2=4\) ✓
\(k=3:\quad F_3=27,\ G_3=16\) ✓
\(k=4:\quad F_4=81,\ G_4=64\) ✓
\(k=5:\quad F_5=243,\ G_5=256\) ✗
So \(k\le4\)
Therefore the maximum \(k\) satisfying both conditions is 4.
Compare
Quantity \(A = 4\)
Quantity \(B = 3\)
Quantity A is greater.
gmatclubot
P, F , and G are three different geometric sequences, where each [#permalink]
11 Sep 2026, 10:32
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