If a, b, c are in AP, and a, b, d are in GP, show that a, (a – b) and (d – c) are in GP.
To prove: a, (a – b) and (d – c) are in GP.
Given: a, b, c are in AP, and a, b, d are in GP
Proof: As a,b,d are in GP then
b2 = ad … (i)
As a, b, c are in AP
2b = (a + c) … (ii)
Considering a, (a – b) and (d – c)
(a – b)2 = a2 – 2ab + b2
= a2 – (2b)a + b2
From eqn. (i) and (ii)
= a2 – (a+c)a + ad
= a2 – a2 - ac + ad
= ad – ac
(a – b)2 = a (d – c)
From the above equation we can say that a, (a – b) and (d – c) are in GP.