If $k + 2, 4 k - 6, 3 k - 2$ are the three consecutive terms of an A.P., then the value of $k$ is

2

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3

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4

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5

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Solution

The correct option is D $3$
It is given that $k + 2, 4 k - 6, 3 k - 2$ are the consecutive terms of A.P, therefore, by arithmetic mean property we have, first term+third term is equal to twice of second term that is:

$(k + 2) + (3 k - 2) = 2 (4 k - 6) \Rightarrow 4 k = 8 k - 12 \Rightarrow 4 k - 8 k = - 12 \Rightarrow - 4 k = - 12 \Rightarrow 4 k = 12 \Rightarrow k = \frac{12}{4} = 3$

Hence $k = 3$ .

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