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The zeroes of $x^{3} - 12 x^{2} + 39 x - k$ are in arithmetic progression. What is $k$ ?

Competition practiceAlgebra

A32
B28
C26
D20

Step-by-Step Solution

Sum

= 12

, so the middle zero is 4:

64 - 192 + 156 - k = 0, k = 28

(zeroes 1, 4, 7).

Answer: B) 28

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