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Question -

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 arein GP.

By using the propertyof 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 arein GP.

By using the propertyof 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 theabove 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 arein GP.

By using the propertyof 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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