Question

plzz answer bhaiyaa log ​

Answer 1

$\boxed{\boxed{\huge{\green{\mathfrak{\underline{Answer}}}}}}$

The numbers 2/3,k,5k/8

are in AP then thier common differen is equal so

consider

a1=2/3

a2=k

a3=5k/8

$\large{a2-a1=a3-a2}$

$\large{k-2/3=5k/8-k}$

$\large{3k-2/3=5k-8k/8}$

$\large{15k-24k=24k-16}$

$\large{15k-24k-24k=16}$

$\large{-33k=16}$

$\boxed{\boxed{\huge{\green{\mathfrak{\underline{k=16/33}}}}}}$

Answer 2

SOLUTION:-

$\frac{2}{3} ,\: k \:, \frac{5k}{8} \: are \: in \: A.P. \\ \\ Here ,\: a = \frac{2}{3} \\ b = k \\ c = \frac{5k}{8}$

Using one of the property of A.P:

If a,b & c are in A.P.

b - a = c - b

$k - \frac{2}{3} = \frac{5k}{8} - k \\ \\ = > k + k - \frac{5k}{8} = \frac{2}{3} \\ \\ = > 2k - \frac{5k}{8} = \frac{2}{ 3} \\ \\ = > \frac{16k - 5k}{8} = \frac{2}{3} \\ \\ = > \frac{11k}{8} = \frac{2}{3} \\ \\ = > 33k = 16 \\ \\ = > k = \frac{16}{33}$

Hope it helps ☺️