Question

If a, b, c are in continued proportion, prove that: `(pa^2+ qab+ rb^2)/(pb^2+qbc+rc^2) = a/c`

Answer 1

Given a, b, c are in continued proportion
`(pa^2+ qab+ rb^2)/(pb^2+qbc+rc^2) = a/c`
Let `a/b = b/c` = k
⇒ a = bk and b = ck     ....(i)
⇒ a = (ck)k = ck²        ...[Using (i)]
and b = ck
L.H.S. = `a/c`
= `(ck^2)/c`
= k²
R.H.S. = `(p(ck^2)^2 + q(ck^2)ck + r(ck)^2)/(p(ck)^2 + q(ck)c + rc^2)`

= `(pc^2k^4 + qc^2k^3 + rc^2k^2)/(pc^2k^2 + qc^2k + rc^2)`

= `(c^2k^2)/(c^2)[(pk^2 + qk + r)/(pk^2 + qk + r)]`
= k²                              ...(iii)
From (ii) and (iii), L.H.s. = R.H.S.
∴ b = ck, a = bk = c k k = ck² 
(i) L.H.S.
= `"a + b"/"b + c"`

= `(ck^2 + ck)/"ck + c"`

= `(ck(k + 1))/(c(k + 1)`
= k
R.H.S.
= `(a^2(b - c))/(b^2(a - b)`

= `((ck^2)^2(ck - c))/((ck)^2(ck^2 - ck)`

= `(c^2k^4c(k - 1))/(c^2k^2(k - 1)`.