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Question
If b is the mean proportion between a and c, show that:
`("a"^4 + "a"^2"b"^2 + "b"^4)/("b"^4 + "b"^2"c"^2 +"c"^4) = "a"^2/"c"^2`
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Solution
Given that b is the mean proportion between a and c.
`"a"/"b" = "b"/"c"` = k
⇒ b = ck ; a = bk = (ck)k = ck2
L.H.S = `("a"^4 + "a"^2"b"^2 + "b"^4)/("b"^4 + "b"^2"c"^2 +"c"^4)`
= `(("ck"^2)^4 + ("ck"^2)^2 ("ck")^2 + ("ck")^4)/(("ck")^4 + (ck)^2"c"^2 + "c"^4)`
= `("c"^4"k"^8 + "c"^4"k"^6 + "c"^4"k"^4)/("c"^4"k"^4 + "c"^4"k"^2 + "c"^4)`
= `("c"^4"k"^4("k"^4 + "k"^2 + 1))/("c"^4("k"^4 + "k"^2 + 1)`
= k4 ..........(i)
R.H.S = `"a"^2/"c"^2 = ("ck"^2)^2/"c"^2 = ("c"^2"k"^4)/"c"^2 = "k"^4`.......(ii)
From (i) and (ii) we get
L.H.S = R.H.S
⇒ `("a"^4 + "a"^2"b"^2 + "b"^4)/("b"^4 + "b"^2"c"^2 +"c"^4) = "a"^2/"c"^2`
Hence proved.
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