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

Given: a, b, c are in GP

From the above, we can have the following conclusion
Formula used: When a,b,c are in GP, b2 = ac Proof: When a,b,c,d are in GP then

⇒ bc = ad … (i)

⇒ b2 = ac … (ii)

⇒ c2 = bd … (iii)

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

(a2 – b2) (c2 – d2) = 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