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, c^{2} = 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., c^{2} = 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+c^{2}

c^{2}=ae

Therefore, a, c, and e are in G.P.

Answered by Pragya Singh | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).