Find the value of $k$ , for which $2 k + 7, 6 k - 2$ and $8 k + 4$ are $3$ consecutive terms of an AP.

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Solution

If three numbers $a, b, c$ are in A.P.

Then $b = \frac{a + c}{2}$ ( $b$ is arithmetic mean of $a$ and $c$ )

If $2 k + 7, 6 k - 2$ and $8 k + 4$ are in A.P.
then,

$6 k - 2 = \frac{2 k + 7 + 8 k + 4}{2}$

$12 k - 4 = 10 k + 11$

$2 k = 15$

$k = \frac{15}{2}$

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