Q17 of 110 Page 41

If a, b, c are in A.P. then prove that (a – c)2 = 4 (b2 – ac).

Given, a, b and c are in A.P.


We know that when t1, t2, t3 … are in A.P., t3 – t2 = t2 – t1


⇒ c – b = b – a


⇒ 2b = a + c


Squaring on both sides,


⇒ (2b)2 = (a + c)2


We know that (a + b)2 = a2 + 2ab + b2.


⇒ 4b2 = a2 + 2ac + c2


Subtracting 4ac on both sides,


⇒ 4b2 – 4ac = a2 + 2ac + c2– 4ac


⇒ 4 (b2 – ac) = a2 – 2ac + c2


∴ 4 (b2 – ac) = (a – c)2


Hence proved.


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