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Question
If a, b, c and dare in continued proportion, then prove that
(a+ d)(b+ c)-(a+ c)(b+ d)= (b-c)2
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Solution
`"a"/"b" = "b"/"c" = "c"/"d" = "k"`
⇒ c = kd
b =kc= k2d
a= kb= k3d
(a+ d)(b + c)-(a+ c)(b + d) = (b- c)2
LHS
(a+ d)(b + c)-(a+ c)(b + d)
= ab + bd + ac + cd - ab - bc - ad - cd
= bd+ ca - bc - ad
= k2d2 + k4d4 - k3d2 - k3d2
= k2d2 + k4d4 - 2k3d2
= k2d2 (1+ k2 - 2k)
RHS
(b - c)2 = (b - c)(b - c)
= b2 - 2bc + c2
= k4d4 - 2k3d2 + k2d2
= k2d2 (k2 - 2k + 1)
LHS = RHS. Hence, proved.
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