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"`

Answer 1

`"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)`