Question

If a, b, care in continued proportion, then show that `(a + b)^2/(ab) = (b + c)^2/(bc)`.

Since a, b, c are in continued proportion

∴ `a/b = square/square` = k(say)

⇒ b = `square`, a = `square` = `square`.k = `square`.k²

Now, L.H.S. = `(a + b)^2/(a.square) = (square + square)^2/(square*square)`

= `(squarek^2(k + 1)^2)/(square*square)`

= `(k + 1)^2/square`

R.H.S. = `(b + c)^2/(b.square) = (square + square)^2/(square*square) = (square (k + 1)^2)/(square*square)`

= `(k + 1)^2/square`

L.H.S. = `square`

Answer 1

Since a, b, c are in continued proportion

∴ `a/b = bb b/bbc` = k(say)

⇒ b = ck, a = bk = ck.k = c.k²

Now, L.H.S. = `(a + b)^2/(a.bb b) = (bb(ck^2) + bb(ck))^2/(bb(ck^2)*bb(ck))`

= `(bb(c^2)k^2(k + 1)^2)/(bb(c^2)*bb(k^3))`

= `(k + 1)^2/bbk`

R.H.S. = `(b + c)^2/(b.bbc) = (bb(ck) + bbc)^2/(bb(ck)*bbc) = (bb(c^2) (k + 1)^2)/(bb(c^2)*bbk)`

= `(k + 1)^2/bbk`

L.H.S. = R.H.S

Hence proved