Question Number 100570 by bshahid010@gmail.com last updated on 27/Jun/20

Commented byDwaipayan Shikari last updated on 27/Jun/20

a^x =b^y =c^z =d^w =k  (k≠0)  a=k^(1/x)   b=k^(1/y)   c=k^(1/z)   d=k^(1/w)   (a/b)=(b/c)=(c/d)    (As they are in G.P)  ad=bc  so      k^((1/x)+(1/w)) =k^((1/y)+(1/z)) ⇒(1/x)+(1/w)=(1/y)+(1/(z ))  or    ac=b^2       k^((1/x)+(1/z)) =k^(2/(y ))    ⇒(1/x)+(1/z)=(2/y)  or  c^2 =bd    k^((1/y)+(1/w)) =k^(2/(z  ))   ⇒(1/y)+(1/w)=(2/z)               They are in H.P