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Question
If a, b, c, d are in continued proportion, prove that (b − c)2 + (c − a)2 + (d − b)2 = (d − a)2.
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Solution
Since a, b, c, d are in continued proportion, we have
`a/b = b/c = c/d` = K (say)
∴ c = dK, b = cK = dK2 and a = bK = dK3.
L.H.S.
= (b − c)2 + (c − a)2 + (d − b)2
= (dK2 − dK)2 + (dK − dK3)2 + (d − dK2)2
= d2K2(K − 1)2 + d2K2 (1 − K2)2 + d2 (1 − K2)2
= d2 [K2(K − 1)2 + K2(K2 − 1)2 + d2(k2 − 1)2]
= d2 [K2 (K − 1)2 + K2 (K − 1)2 (K + 1)2 + (K − 1)2 (K + 1)2]
= d2 (K − 1)2 [K2 + K2 (K+ 1)2 + (K + 1)2]
= d2 (K − 1)2 [K2 + K2 (K2 + 2K + 1) + K2 + 2K + 1]
= d2 (k − 1)2 [K4 + 2K3 + 3K2 + 2K + 1]
= d2(K − 1)2 (K2 + K + 1)2
= d2[(K − 1) (K2 + K + 1)2]
= d2 (K3 − 1)2 = (dK3 − d)2 = (a − d)2 = (d − a)2 = R.H.S.
Hence, (b − c)2 + (c − a)2 + (d − b)2 = (d − a)2.
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