1.

If a, b, c, d are in G.P., prove that:(i) (ab – cd) / (b2 – c2) = (a + c) / b(ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2(iii) (b + c) (b + d) = (c + a) (c + d)

Answer»

(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.


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