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 G.P.

Given a, b, c are in A.P.

Hence, b – a = c – b  … (1)

And, given that b, c, d are in G.P.

So, c² = bd  ..........… (2)

Also, \( \dfrac{1}{c}\), \( \dfrac{1}{d}\), \( \dfrac{1}{e}\) are in A.P.

So,

\( \dfrac{1}{d}- \dfrac{1}{c}= \dfrac{1}{e}- \dfrac{1}{d}\)

\( \dfrac{2}{d}= \dfrac{1}{c}+ \dfrac{1}{e}\) ................(3)

Now, required to prove that a, c, e are in G.P. i.e., c² = ae

From (1), we have

2b = a + c

b =\( \dfrac{(a+c)}{2}\)

And from (2), we have

d = \(\dfrac{c^2}{b}\)

On substituting these values in (3), we get

= \( \dfrac{2b}{c^2}=\dfrac{1}{c}+\dfrac{1}{e}\)

= \( \dfrac{2(a+c)}{2c^2}=\dfrac{1}{c}+\dfrac{1}{e}\)

 = \( \dfrac{a+c}{c^2}=\dfrac{e+c}{ce}\)

= \( \dfrac{a+c}{c}=\dfrac{e+c}{e}\)

= (a+c)e=(e+c)c

= ae+ce=ec+c²

c²=ae

Therefore, a, c, and e are in G.P.