If a, b, c are in G.P., prove that :
(a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2.
As,
a, b, c are in G.P, let r be the common ratio.
Therefore,
b = ar … (1)
c = ar2 … (2)
To prove: (ab + bc + cd)2 = (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2
As, LHS = (a + 2b + 2c) (a – 2b + 2c)
⇒ LHS = (a + 2ar + 2ar2)(a – 2ar + 2ar2)
⇒ LHS = a2(1 + 2r + 2r2)(1 – 2r + 2r2)
⇒ LHS = a2 (1 + 4r2 + 4r4 – 4r2)
⇒ LHS = a2(1 + 4r4)
And RHS = a2 + 4a2r4 = a2(1 + 4r4)
Clearly, LHS = RHS
Hence proved