Answer :

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.


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