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Q.

For what value of k, $k + 2$ , $4 k - 6$ , $3 k - 2$ are three consecutive terms of an A.P ?

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a

–3

b

1

c

3

d

–1

answer is C.

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Detailed Solution

k + 2

,

4 k - 6

,

3 k - 2

are in A.P

$\Rightarrow 2 (4 k - 6) = k + 2 + 3 k - 2$

$\Rightarrow 8 k - 12 = 4 k$

$\Rightarrow 4 k = 12 \Rightarrow k = 3$ .

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