Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
Let us consider LHS: \( a^2b^2c^2 [\dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3}]\)
\(\dfrac{ (ac)c^2}{a} + \dfrac{(b^2)^2}{b} + \dfrac{a^2(ac)}{c}\) [by substituting the b2 = ac]
\( \dfrac{ac^3}{a} + \dfrac{b^4}{b} +\dfrac{ a^3c}{c}\)
c3 + b3 + a3 = RHS
LHS = RHS
Hence proved.
Answered by Aaryan | 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) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).