Find the value of K so that $(K + 2), (4 K - 6)$ and $(3 K - 6)$ are three continuous terms of AP.

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Solution

Given $(K + 2), (4 K - 6)$ and $(3 K - 6)$ are three continuous terms of AP

As the difference between any two consecutive terms of A.P is constant.

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

$\Rightarrow 3 k - 8 = - k$

$\Rightarrow 4 k = 8$

$\Rightarrow k = 2$ .

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