Question

If a, b, c and d are in G.P. show that (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)² .

Answer 1

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

It has to be proved that,

(a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

R.H.S.

= (ab + bc + cd)²

= (ab + ad + cd)² [Using (1)]

= [ab + d (a + c)]²

= a²b² + 2abd (a + c) + d² (a + c)²

= a²b² +2a²bd + 2acbd + d²(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c² [Using (1) and (2)]

= a²b² + a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c²

= a²b² + a²c² + a²d² + b² × b² + b²c² + b²d² + c²b² + c² × c² + c²d²

[Using (2) and (3) and rearranging terms]

= a²(b² + c² + d²) + b² (b² + c² + d²) + c² (b²+ c² + d²)

= (a² + b² + c²) (b² + c² + d²)

= L.H.S.

∴ L.H.S. = R.H.S.