If a, b, c, d are in G.P, prove that : (a + b + c + d)2 = (a + b)2 +

1.	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
Answer» 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.

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

Answer»

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.