(i) (ab – cd) / (b2 – c2) = (a + c) / b
(ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2
Solution:
(i) (ab – cd) / (b2 – c2) = (a + c) / b
According to the question a, b, c are in GP.
Making use of the property of geometric mean, we can write
b2 = ac
bc = ad
c2 = bd
Let us take the LHS: (ab – cd) / (b2 – c2)
$ =\left( ab-cd \right)\text{ }/\text{ }\left( {{b}^{2}}-{{c}^{2}} \right)\text{ } $
$ =\text{ }\left( ab-cd \right)\text{ }/\text{ }\left( ac-bd \right) $
$ =\text{ }\left( ab-cd \right)b\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left( a{{b}^{2}}-bcd \right)\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left[ a\left( ac \right)-c\left( {{c}^{2}} \right) \right]\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left( {{a}^{2}}c-{{c}^{3}} \right)\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left[ c\left( {{a}^{2}}-{{c}^{2}} \right) \right]\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left[ \left( a+c \right)\text{ }\left( ac-{{c}^{2}} \right) \right]\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left[ \left( a+c \right)\text{ }\left( ac-bd \right) \right]\text{ }/\text{ }\left( ac-bd \right)b $
$ =\text{ }\left( a+c \right)\text{ }/\text{ }b $
$ =\text{ }RHS $
∴ LHS = RHS
Hence proved.
(ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2
According to the question a, b, c are in GP.
Making use of the property of geometric mean, we can write
b2 = ac
bc = ad
c2 = bd
Let us consider RHS: (a + b)2 + 2(b + c)2 + (c + d)2
Let us expand
$ ={{\left( a\text{ }+\text{ }b \right)}^{2}}~+\text{ }2{{\left( b\text{ }+\text{ }c \right)}^{2}}~+\text{ }{{\left( c\text{ }+\text{ }d \right)}^{2}}~ $
$ =\text{ }{{\left( a\text{ }+\text{ }b \right)}^{2}}~+\text{ }2\text{ }\left( a+b \right)\text{ }\left( c+d \right)\text{ }+\text{ }{{\left( c+d \right)}^{2}} $
$ =\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2ab\text{ }+\text{ }2\left( {{c}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2cb \right)\text{ }+\text{ }{{c}^{2}}~+\text{ }{{d}^{2}}~+\text{ }2cd $
$ =\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }{{c}^{2}}~+\text{ }{{d}^{2}}~+\text{ }2ab\text{ }+\text{ }2\left( {{c}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2cb \right)\text{ }+\text{ }2cd $
$ =\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }{{c}^{2}}~+\text{ }{{d}^{2}}~+\text{ }2\left( ab\text{ }+\text{ }bd\text{ }+\text{ }ac\text{ }+\text{ }cb\text{ }+cd \right)\text{ }\left[ Since,\text{ }{{c}^{2}}~=\text{ }bd,\text{ }{{b}^{2}}~=\text{ }ac \right] $
We can easily understand 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.