Determine $k$ so that $k + 2, 4 k - 6 a n d 3 k - 2$ are three consecutive terms of A.P.

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Solution

As it is given that $k + 2, 4 k - 6 and 3 k - 2$ are the consecutive terms of A.P,
By the arithmetic mean property, first term + third term is equal to twice of the second term that is:
If $a, b$ and $c$ are in AP then $2 b = a + c$
So, we get,
$(k + 2) + (3 k - 2) = 2 (4 k - 6) \Rightarrow 4 k = 8 k - 12 \Rightarrow 4 k - 8 k = - 12 \Rightarrow - 4 k = - 12 \Rightarrow 4 k = 12 \Rightarrow k = 12 / 4 \Rightarrow k = 3$
Hence value of $K$ is $3$

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