Determine k, so that k+2, 4k-6 and 3k-2 are t | vedclass

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(D) Since $k+2, 4k-6$ and $3k-2$ are in $A.P.$,the common difference between consecutive terms must be equal.Therefore,$(4k-6) - (k+2) = (3k-2) - (4k-

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$3$

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(D) Since $k+2, 4k-6$ and $3k-2$ are in $A.P.$,the common difference between consecutive terms must be equal.Therefore,$(4k-6) - (k+2) = (3k-2) - (4k-6)$.Simplifying the left side: $4k - 6 - k - 2 = 3k - 8$.Simplifying the right side: $3k - 2 - 4k + 6 = -k + 4$.Equating both sides: $3k - 8 = -k + 4$.Adding $k$ to both sides: $4k - 8 = 4$.Adding $8$ to both sides: $4k = 12$.Dividing by $4$: $k = 3$.

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

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