Given, a, b, c,and d are in G.P.

So, we have

b^{2} = ac … (i)

c^{2} = bd … (ii)

ad = bc … (iii)

Required to prove (a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n}) are in G.P. i.e.,

(b^{n} + c^{n})^{2} = (a^{n} + b^{n}) (c^{n} + d^{n})

Taking L.H.S.

(b^{n} + c^{n})^{2} = b^{2n }+ 2b^{n}c^{n} + c^{2n}

= (b^{2})^{n}+ 2b^{n}c^{n} + (c^{2})^{ n}

= (ac)^{n} + 2b^{n}c^{n} + (bd)^{n} [Using (i) and (ii)]

= a^{n} c^{n} + b^{n}c^{n}+ b^{n} c^{n} + b^{n} d^{n}

= a^{n} c^{n} + b^{n}c^{n}+ a^{n} d^{n} + b^{n} d^{n} [Using (iii)]

= c^{n} (a^{n} + b^{n}) + d^{n} (a^{n} + b^{n})

= (a^{n} + b^{n}) (c^{n} + d^{n})

= R.H.S.

Therefore, (a^{n} + b^{n}), (b^{n} + c^{n}), and (c^{n} + d^{n}) are in G.P

Hence proved.

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 \( \dfrac{81}{2}\).