Question

Determine k so that 2/3,k and 5/8k are the three consecutive terms of an a.p

Answer 1

Solution is attached below

Answer 2

Answer:

$Value \: k = \frac{16}{33}$

Step-by-step explanation:

$Given,\\\frac{2}{3},k\: and \: \frac{5k}{8}\\are \: three \: consecutive\:terms \\of\:an\:A.P$

$We\:know\:that,\\If\:a,b\:and \:c \: are \:in \:A.P\\then \: b-a = c-b$

$\implies k-\frac{2}{3}=\frac{5k}{8}-k$

$\implies \frac{3k-2}{3}=\frac{5k-8k}{8}$

$\implies \frac{3k-2}{3}=\frac{-3k}{8}$

$\implies 8(3k-2)=3(-3k)$

$\implies 24k-16=-9k$

$\implies 24k+9k = 16$

$\implies 33k = 16$

$\implies k = \frac{16}{33}$

Therefore,

$Value \: k = \frac{16}{33}$

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