Guest

if a b c are in gp and a^1/x=b^1/y=c^1/z prove that x y z are in ap

if a b c are in gp and a^1/x=b^1/y=c^1/z prove that x y z are in ap

Grade:11

3 Answers

Arun
25750 Points
6 years ago
z , c = ky, b = kxa = k k (let) =z = c1/y = b1/x1/a
now a,b,c are in g.p. (given)
hence b/a = c/b
 
ky / kx  = kz / ky
k(2y) = k(x +z)
hence 2y = x + z
hence x, y, z will be in A.P.
Mohd Mujtaba
131 Points
6 years ago
Let a^1/x=b^1/y=c^1/z=m .soa=m^x,b=m^y, c=m^z. Now a b c are in gp so b^2=ac. This imply m^2y=m^z.m^x so comparing we get 2y=x+z. Hence a b c are in ap thank☺☺☺☺
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your question
 
Let,
a1/x = b1/y = c1/z = k
Hence,
a = kx, b = ky, c = kz
Since a, b, c are in GP
b2 = ac
k2y = k(x+z)
or, 2y = x + z
Hence, x, y, z are in A.P.
 
Hope it helps.
Thanks and regards,
Kushagra
 

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free