If a, b, c, d are in G.P, prove that :
(a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2
a, b, c, d are in G.P.
Therefore,
bc = ad … (1)
b2 = ac … (2)
c2 = bd … (3)
If somehow we use RHS and Make it equal to LHS, our job will be done.
we can manipulate the RHS of the given equation as –
Note: Here we are manipulating RHS because working with a simpler algebraic equation is easier and this time RHS is looking simpler.
RHS = (a + b)2 + 2(b + c)2 + (c + d)2
⇒ RHS = a2 + b2 + 2ab + 2(c2 + b2 + 2cb) + c2 + d2 + 2cd
⇒ RHS = a2 + b2 + c2 + d2 + 2ab + 2(c2 + b2 + 2cb) + 2cd
Put c2 = bd and b2 = ac, we get –
⇒ RHS = a2 + b2 + c2 + d2 + 2(ab + ad + ac + cb + cd)
You can visualize the above expression by making separate terms for (a + b + c)2 + d2 + 2d(a + b + c) = {(a + b + c) + d}2
⇒ RHS = (a + b + c + d)2 = LHS
Hence Proved.