# 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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