Question

Find
the value of k so that
2/3 ,K, 5/3k
are three consecutive terms of an A.P​

Answer 1

Answer:

given terms of AP are , : ⅔, k, 5k/3

we know that if a,b,c are in ap

then, 2b = a+ c

$2k = \frac{2}{3} + \frac{5k}{3} \\2k = \frac{2 + 5k}{3} \\ = > 6k = 2 + 5k \\ = > 6k - 5k = 2 \\ = > k = 2$

so the value of k is 2.