Question

If a, b, c, d are in GP, prove that (a² – b²), (b² – c²), (c² – d²) are in GP.

Answer 1

To prove: (a² – b² ), (b² – c² ), (c² – d² ) are in GP. 

Given: a, b, c are in GP 

Formula used: When a,b,c are in GP, b² = ac 

Proof: When a,b,c,d are in GP then

From the above, we can have the following conclusion 

⇒ bc = ad … (i) 

⇒ b² = ac … (ii) 

⇒ c² = bd … (iii) 

Considering (a² – b² ), (b² – c² ), (c² – d² ) (a² – b² ) (c² – d ² ) 

= a²c² – a²d² – b²c² + b²d² 

= (ac)² – (ad)² – (bc)² + (bd)² 

From eqn. (i) , (ii) and (iii)

= (b²)² – (bc)² – (bc)² + (c²)² 

= b⁴ – 2b²c² + c⁴ (a² – b² ) (c² – d²) 

= (b² – c²)² 

From the above equation we can say that (a² – b² ), (b² – c² ), (c² – d² ) are in GP