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g(n)=(-1)^n \time c \time n
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17 Aug 2026, 10:12
17 Aug 2026, 10:12
1
Expert Reply
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Question Stats:
50% (03:17) correct
50% (04:42) wrong based on 4 sessions
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\(g(n)=(-1)^n \times c \times n\) if n is even, and \(g(n)=(-1)^n \times d \times n\). if n is odd, where c and d are positive constants. The difference between the largest and smallest among \(g(1), g(2), g(3) \) is 20 , and the difference between the largest and smallest among \(g(2), g(3), g(4)\) is 34 .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(g(6)\) |
\(42\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
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
g(n)=(-1)^n \time c \time n
[#permalink]
26 Aug 2026, 03:53
26 Aug 2026, 03:53
1
Expert Reply
Step 1: Evaluate g(n) for individual values of $n$
Using the definition:
\(g(n)=c \cdot n\) when n is even (since \((-1)^n=1\) )
\(g(n)=-d \cdot n \) when n is odd (since \((-1)^n=-1\) )
With positive constants c>0 and d>0 :
\(g(1)=(-1)^1 \cdot d \cdot 1=-d \)
\( g(2)=(-1)^2 \cdot c \cdot 2=2 c \)
\( g(3)=(-1)^3 \cdot d \cdot 3=-3 d \)
\(g(4)=(-1)^4 \cdot c \cdot 4=4 c \)
Step 2: Analyze the first given range \(\{g(1), g(2), g(3)\}\)
The set of values is \(\{-d, 2 c,-3 d\} \).
Since c>0 and d>0 , the largest value is 2 c (positive) and the smallest value is (most negative).
The difference between the largest and smallest is 20:
\(2 c-(-3 d)=20 \Longrightarrow 2 c+3 d=20 \quad-(\text { Equation } 1)\)
Step 3: Analyze the second given range \(\{g(2), g(3), g(4)\}\)
The set of values is \( \{2 c,-3 d, 4 c\}\) .
The largest value is 4 c (since \(4 c>2 c>0\) ) and the smallest value is -3 d .
The difference between the largest and smallest is 34 :
\(4 c-(-3 d)=34 \Longrightarrow 4 c+3 d=34 \quad \longrightarrow \text { (Equation 2) }\)
Step 4: Solve for c and d
Subtract Equation 1 from Equation 2:
\((4 c+3 d)-(2 c+3 d)=34-20 \\
2 c=14 \Longrightarrow c=7\)
Substitute c=7 into Equation 1:
\(2(7)+3 d=20 \Longrightarrow 14+3 d=20 \Longrightarrow 3 d=6 \Longrightarrow d=2\)
Step 5: Calculate Quantity A
Since n=6 is even:
\(g(6)=(-1)^6 \cdot c \cdot 6=1 \cdot 7 \cdot 6=42\)
Conclusion
Quantity A: g(6)=42
Quantity B: 42
The two quantities are equal.
Using the definition:
\(g(n)=c \cdot n\) when n is even (since \((-1)^n=1\) )
\(g(n)=-d \cdot n \) when n is odd (since \((-1)^n=-1\) )
With positive constants c>0 and d>0 :
\(g(1)=(-1)^1 \cdot d \cdot 1=-d \)
\( g(2)=(-1)^2 \cdot c \cdot 2=2 c \)
\( g(3)=(-1)^3 \cdot d \cdot 3=-3 d \)
\(g(4)=(-1)^4 \cdot c \cdot 4=4 c \)
Step 2: Analyze the first given range \(\{g(1), g(2), g(3)\}\)
The set of values is \(\{-d, 2 c,-3 d\} \).
Since c>0 and d>0 , the largest value is 2 c (positive) and the smallest value is (most negative).
The difference between the largest and smallest is 20:
\(2 c-(-3 d)=20 \Longrightarrow 2 c+3 d=20 \quad-(\text { Equation } 1)\)
Step 3: Analyze the second given range \(\{g(2), g(3), g(4)\}\)
The set of values is \( \{2 c,-3 d, 4 c\}\) .
The largest value is 4 c (since \(4 c>2 c>0\) ) and the smallest value is -3 d .
The difference between the largest and smallest is 34 :
\(4 c-(-3 d)=34 \Longrightarrow 4 c+3 d=34 \quad \longrightarrow \text { (Equation 2) }\)
Step 4: Solve for c and d
Subtract Equation 1 from Equation 2:
\((4 c+3 d)-(2 c+3 d)=34-20 \\
2 c=14 \Longrightarrow c=7\)
Substitute c=7 into Equation 1:
\(2(7)+3 d=20 \Longrightarrow 14+3 d=20 \Longrightarrow 3 d=6 \Longrightarrow d=2\)
Step 5: Calculate Quantity A
Since n=6 is even:
\(g(6)=(-1)^6 \cdot c \cdot 6=1 \cdot 7 \cdot 6=42\)
Conclusion
Quantity A: g(6)=42
Quantity B: 42
The two quantities are equal.
gmatclubot
g(n)=(-1)^n \time c \time n [#permalink]
26 Aug 2026, 03:53
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