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Question
If u, v, w, and x are in continued proportion, then prove that (2u+3x) : (3u+4x) : : (2u3+3v3) : (3u3+4v3)
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Solution
`"u"/"v" = "v"/"w" = "w"/"x" = "a"`
w = ax
v = aw = a2x
u = av = a3x
LHS
`(2"u" + 3"x")/(3"u" + 4"x")`
`= (2"a"^3"x" + 3"x")/(3"a"^3"x" + 4"x")`
`= (2"a"^3 + 3)/(3"a"^3 + 4)`
RHS
`(2"u"^3 + 3"v"^3)/(3"u"^3 + 4"v"^3)`
`= (2"a"^9"x"^3 + 3"a"^6"x"^3)/(3"a"^9"x"^3 + 4"a"^6"x"^3)`
`= (2"a"^3 + 3)/(3"a"^3 + 4)`
LHS = RHS. Hence , proved.
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