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
\(\frac{(b-c)^2}{(a-b)^2}=\frac{(c-d)^2}{(b-c)^2}\\ LHS=\frac{b^2+c^2-2bc}{a^2+b^2-2ab}\\ LHS=\frac{ac+c^2-2bc}{a^2+ac-2ab}\\ LHS=\frac{c(a+c-2b)}{a(a+c-2b)}\\ LHS=\frac{c}{a}\\ LHS=\frac{ar^2}{a}\\ LHS=r^2\\\)
\(RHS=\frac{(c-d)^2}{(b-c)^2}\\ RHS=\frac{(c^2+d^2-2cd)}{(b^2+c^2-2bc)}\\ RHS=\frac{(bd+d^2-2cd)}{(b^2+db-2bc)}\\ RHS=\frac{d(b+d-2c)}{b(b+d-2c)}\\ RHS=\frac{d}{b}\\ RHS=\frac{ar^3}{ar}\\ RHS=r^2\\ RHS=LHS\\ \therefore\\ \mbox{The terms in question (i) make a GP}\)
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+118728 
