Find $k$ such that $k + 2, 4 k - 6$ and $3 k - 2$ are three succesive terms of an $A . P$

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Solution

Given,

$k + 2, 4 k - 6, 3 k - 2$ are in A.P

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

$3 k - 8 = - k + 4$

$4 k = 12$

$∴ k = 3$

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Similar questions

Determine k so that (3k -2), (4k-6) and (k +2) are three consecutive terms of an AP.

(K + 2), (4 K - 6)

(3 K - 6)

k

k^{2} + 4 k + 8

2 k^{2} + 3 k + 6

3 k^{2} + 4 k + 4

k

k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6, 3 k^{2} + 4 k + 4

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