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Question
If a, b, c and dare in continued proportion, then prove that
`sqrt (("a + b + c")("b + c + d")) = sqrt "ab" + sqrt "bc" + sqrt "cd"`
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Solution
`"a"/"b" = "b"/"c" = "c"/"d" = "k"`
⇒ c = kd
b =kc= k2d
a= kb= k3d
`sqrt (("a + b + c")("b + c + d")) = sqrt "ab" + sqrt "bc" + sqrt "cd"`
LHS
`sqrt (("a + b + c")("b + c + d"))`
`= sqrt (("k"^3"d" + "k"^2"d" + "kd")("k"^2"d" + "kd" + "d"))`
`= sqrt ("kd" ("k"^2 + "k" + 1) xx "d"("k"^2 + "k" + 1))`
`= sqrt ("kd"^2 ("k"^2 + "k" + 1)^2)`
`= "d" sqrt "k" ("k"^2 + "k" + 1)`
RHS
= `sqrt "ab" + sqrt "bc" + sqrt "cd"`
= `sqrt ("k"^3"d" xx "k"^2"d") + sqrt ("k"^2"d" xx "kd") + sqrt ("kd" xx "d")`
=`sqrt ("k"^5"d"^2) + sqrt ("k"^3"d"^2) + sqrt "k" "d"^2`
= `"k"^2"d" sqrt "k" + "kd" sqrt "k" + "d" sqrt "k"`
= `"d" sqrt "k" ("k"^2 + "k" + 1)`
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