It is given that a,b,c are in A.P., b,c,d are in G.P. and 1 c , 1 d , 1 e are in A.P.
Since a,b,c are in A.P.,
b−a=c−b 2b=a+c b= a+c 2
Also, b,c,d are in G.P.
c 2 =bd d= c 2 b
Also, 1 c , 1 d , 1 e are in A.P.
1 d − 1 c = 1 e − 1 d 2 d = 1 c + 1 e (1)
Substitute value of b and d in (1),
2b c 2 = 1 c + 1 e 2( a+c ) 2 c 2 = 1 c + 1 e a+c c 2 = c+e ce a+c c = e+c e
Further simplify above equation.
( a+c )e=( e+c )c ae+ce=ec+ c 2 c 2 =ae
Thus, a,c,e are in G.P.
If a b and c are in GP and a1/x=b1/y=c1/z, prove that x, y and z are in AP.