Determine k so that frac{2}{3}, k and frac{5} | vedclass

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Question

(A) Since $\frac{2}{3}, k, \frac{5}{8} k$ are in $A.P.$,the common difference between consecutive terms must be equal.Therefore,$k - \frac{2}{3} = \fr

Vedclass · Accepted Answer

$\frac{16}{33}$

Vedclass · Answer

(A) Since $\frac{2}{3}, k, \frac{5}{8} k$ are in $A.P.$,the common difference between consecutive terms must be equal.Therefore,$k - \frac{2}{3} = \frac{5}{8} k - k$.Rearranging the terms to isolate $k$ on one side:$k - \frac{5}{8} k + k = \frac{2}{3}$$2k - \frac{5}{8} k = \frac{2}{3}$Finding a common denominator for the left side:$\frac{16k - 5k}{8} = \frac{2}{3}$$\frac{11k}{8} = \frac{2}{3}$Solving for $k$:$k = \frac{2}{3} 	imes \frac{8}{11} = \frac{16}{33}$.

Determine $k$ so that $\frac{2}{3}, k$ and $\frac{5}{8} k$ are the three consecutive terms of an $A.P.$

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