(i) (ab – cd) / (b2 – c2) = (a + c) / b
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
bc = ad
c2 = bd
Let us consider LHS: (ab – cd) / (b2 – c2)
(ab – cd) / (b2 – c2) = (ab – cd) / (ac – bd)
= (ab – cd)b / (ac – bd)b
= (ab2 – bcd) / (ac – bd)b
= [a(ac) – c(c2)] / (ac – bd)b
= (a2c – c3) / (ac – bd)b
= [c(a2 – c2)] / (ac – bd)b
= [(a+c) (ac – c2)] / (ac – bd)b
= [(a+c) (ac – bd)] / (ac – bd)b
= (a+c) / b
= RHS
∴ LHS = RHS
Hence proved.
(ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
bc = ad
c2 = bd
Let us consider RHS: (a + b)2 + 2(b + c)2 + (c + d)2
Let us expand
(a + b)2 + 2(b + c)2 + (c + d)2 = (a + b)2 + 2 (a+b) (c+d) + (c+d)2
= a2 + b2 + 2ab + 2(c2 + b2 + 2cb) + c2 + d2 + 2cd
= a2 + b2 + c2 + d2 + 2ab + 2(c2 + b2 + 2cb) + 2cd
= a2 + b2 + c2 + d2 + 2(ab + bd + ac + cb +cd) [Since, c2 = bd, b2 = ac]
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 = LHS
Hence proved.
(iii) (b + c) (b + d) = (c + a) (c + d)
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
bc = ad
c2 = bd
Let us consider LHS: (b + c) (b + d)
Upon expansion we get,
(b + c) (b + d) = b2 + bd + cb + cd
= ac + c2 + ad + cd [by using property of geometric mean]
= c (a + c) + d (a + c)
= (a + c) (c + d)
= RHS
∴ LHS = RHS
Hence proved.