Question:
If the variance of the numbers \( 9, 15, 21, \dots, (6n+3) \) is P, then the variance of the first \( n \) even numbers is
If the variance of the numbers \( 9, 15, 21, \dots, (6n+3) \) is P, then the variance of the first \( n \) even numbers is
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\( \text{Var}(aX+b) = a^2 \text{Var}(X) \).
Updated On: May 16, 2026
- 9P
- 3P
- P/9
- P/3
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The Correct Option is C
Solution and Explanation
Step 1: Analyze Series:
Series 1: \( 6k+3 \). Variance \( P = \text{Var}(6k+3) = 6^2 \text{Var}(k) = 36 \text{Var}(k) \). Series 2: \( 2k \) (First \( n \) even numbers). Variance \( V' = \text{Var}(2k) = 2^2 \text{Var}(k) = 4 \text{Var}(k) \).
Step 2: Relate Variances:
From first equation, \( \text{Var}(k) = P/36 \). Substitute into second equation: \( V' = 4(P/36) = P/9 \).
Step 3: Final Answer:
P/9.
Series 1: \( 6k+3 \). Variance \( P = \text{Var}(6k+3) = 6^2 \text{Var}(k) = 36 \text{Var}(k) \). Series 2: \( 2k \) (First \( n \) even numbers). Variance \( V' = \text{Var}(2k) = 2^2 \text{Var}(k) = 4 \text{Var}(k) \).
Step 2: Relate Variances:
From first equation, \( \text{Var}(k) = P/36 \). Substitute into second equation: \( V' = 4(P/36) = P/9 \).
Step 3: Final Answer:
P/9.
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