If 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. prove that a,c,e are in GP.

Explanation

It is given that a,b,c are in A.P. Thus,
2b=a+c (1)
It is given that b,c,d are in G.P. Thus
`c^(2)=bd`
Also,`1/c,1/d,1/e` are in A.P
So, `2/d=1/c+1/e` (3)
It has proved that a,c,e are in G.P. i.e., `c^(2)=ae`
From (1), we get
`b=(a+c)/2`
From (2), we get
`d=(c^(2))/b`
Substituting these values in (3), we get
`(2b)/(c^(2)==1/c+1/e`
or `(a+c)/(c^(2))=(e+c)/(ce)`
or `(a+c)/c=(e+c)/e`
or `ae+ce=ec+c^(2)`
or `c^(2)=ae`
Thus, a,c, and e are in G.P.