Question

if 1+k,5/6+k,13/18+k are in GP value of k is

Answer 1

Given:-- 1+k,5/6+k & 13/18+k are in G.P

therefore,according to the property of G.P...we have

[(1+k)(13/18+k)]^1/2=5/6+k

=(1+k)(13/18+k)=[5/6+k]^2

=(1+k)(13+18k)/18=(25+36k^2+60k)/36

=2(13+31k+18k^2)=36k^2+60k+25

=2k+1=0

that is...k=-1/2.

Answer 2

Geometric Progression

$1+k,\frac{5}{6}+k, \frac{13}{18} +k$ are in GP, this means ,

$\frac{(\frac{5}{6}+k )}{(1+k)}=\frac{(\frac{13}{18}+k) }{(\frac{5}{6}+k) }$

On solving this equation, we get,

$(\frac{5}{6}+k)^2 =(1+k)(\frac{13}{18}+k)$

$\implies (\frac{5}{6})^2+k^2+2\times\frac{5}{6}\times k=\frac{13}{18}+\frac{13k}{18}+k+k^2$

$\implies \frac{25}{36}+k^2+\frac{5k}{3}=\frac{13}{18}+\frac{31k}{18}+k^2\\ \\ \implies \frac{-1}{36}=\frac{k}{18}\\ \\ \implies k=-2$

Hence the value of k is -2.