Question

If a, b, c, d are in continued proportion, prove that:

`((a -b)/c + (a - c)/b)^2 - ((d - b)/c + (d - c)/b)^2 = (a - d)^2 (1/c^2 - 1/b^2)`

Answer 1

a, b, c, d are in continued proportion

∴ `a/b = b/c = c/d` = k(say)

∴ c = dk, b = ck = dk, k = dk²

a = bk = dk². k = dk³

L.H.S. = `((a -b)/c + (a - c)/b)^2 - ((d - b)/c + (d - c)/b)^2`

= `((dk^3 - dk^2)/(dk) + (dk^3 - dk)/(dk^2))^2 - ((d - dk^2)/(dk) + (d - dk)/(dk^2))^2`

= `((dk^2(k - 1))/(dk) + (dk(k^2 - 1))/(dk^2))^2 - ((d(1 - k^2))/(dk) + (d( 1 - k^2))/(dk^2))^2`

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

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

= `((k^3 - 1)^2)/k^2 - (-k^3 + 1)^2/k^4`

= `(k^3 - 1)^2/k^2 - (1 - k^3)^2/k^4`

= `((k^3 - 1)/k^2)^2 ((1 - 1)/k^2)`

= `((k^3 - 1)^2(k^2 - 1))/k^4`

= `((k^3 - 1)^2(k^2 - 1))/k^4`

R.H.S. = `(a  –  d)^2(1 / c^2 - 1/b^2)`

= `(dk^3 - d)^2(1 / (d^2k^2) - (1)/(d^2k^4))`

= `d^2(k^3 - 1)^2((k^2 - 1)/(d^2k^4))`

= `((k^3 - 1)^2(k^2 - 1))/k^4`
∴ L.H.S. = R.H.S.