If a,b,c,d are in G.P ,then prove that-

Here are answers to the first three.  Let's see if you can make use of these to do part (iv) yourself:

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Hi AaratikRoy :)

If a,b,c,d are in G.P ,then prove that-

(i) (a-b)^2 , (b-c)^2 , (c-d)^2 are in G.P 

If a, b, c, d are in GP then there there is a real number r such that

$a\\ b=ar\\ c=ar^2\\ d=ar^3\\$

also

$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\\ so\\ b^2=ac\qquad c^2=bd$

Now I will look at the sequence in question (i)

(i) (a-b)^2 , (b-c)^2 , (c-d)^2 are in G.P 

This sequence is a GP if it can be shown that

$\frac{(b-c)^2}{(a-b)^2}=\frac{(c-d)^2}{(b-c)^2}\\ LHS=\frac{b^2+c^2-2bc}{a^2+b^2-2ab}\\ LHS=\frac{ac+c^2-2bc}{a^2+ac-2ab}\\ LHS=\frac{c(a+c-2b)}{a(a+c-2b)}\\ LHS=\frac{c}{a}\\ LHS=\frac{ar^2}{a}\\ LHS=r^2\\$

$RHS=\frac{(c-d)^2}{(b-c)^2}\\ RHS=\frac{(c^2+d^2-2cd)}{(b^2+c^2-2bc)}\\ RHS=\frac{(bd+d^2-2cd)}{(b^2+db-2bc)}\\ RHS=\frac{d(b+d-2c)}{b(b+d-2c)}\\ RHS=\frac{d}{b}\\ RHS=\frac{ar^3}{ar}\\ RHS=r^2\\ RHS=LHS\\ \therefore\\ \mbox{The terms in question (i) make a GP}$

(iv) (b-c)^2+(c-a)^2+(d-b)^2= (a-d)^2

I think you just want to show that this is true........???

a 

b = ar

c = ar^2

d = ar^3

So

(b - c)^2   = (ar - ar^2)^2  = [ar (1 - r)]^2  =  (ar)^2 (1 - r)^2  = (ar)^2 (1 - r)^2

(c - a)^2  =  (ar^2 - a)^2  =  [ a (r^2 - 1)]^2  = (a)^2 (r + 1)^2 (1 - r)^2

(d - b)^2  = ( ar^3 - ar)^2  = [ ar ( r^2 -1)]^2 = (ar)^2 (r + 1)^2 (1 - r)^2

(a - d)^2  = (a - ar^3)^2  = [  a (1 - r^3)]^2 = (a)^2  (1 - r)^2 (  1 + r + r^2)^2

So

(b-c)^2+(c-a)^2+(d-b)^2= (a-d)^2

(ar)^2 (1 - r)^2  + (a)^2 (r + 1)^2 (1 - r)^2 + (ar)^2 (r + 1)^2 (1 - r)^2  = (a)^2  (1 - r)^2 (  1 + r + r^2)^2

Divide through by   a(1 - r)^2

r^2  + (r + 1)^2   + r^2 (1 + r)^2   =  ( 1 + r + r^2) ^2

r^2 + [ r^2 + 2r + 1] + r^2 ( r^2 + 2r + 1) =

2r^2 + 2r + 1  + r^4 + 2r^3 + r^2  =

r^4 + 2r^3 + 3r^2 + 2r + 1 =

This factors as :

( 1 + r + r^2)^2

And this  = the RHS