Question

If a, b, c are in continued proportion, prove that: ${\displaystyle \frac{p{a}^2+qab+r{b}^2}{p{b}^2+qbc+r{c}^2}=\frac{a}{c}}$

Answer 1

Given a, b, c are in continued proportion
${\displaystyle \frac{p{a}^2+qab+r{b}^2}{p{b}^2+qbc+r{c}^2}=\frac{a}{c}}$
Let ${\displaystyle \frac{a}{b}=\frac{b}{c}}$ = k
⇒ a = bk and b = ck     ....(i)
⇒ a = (ck)k = ck²        ...[Using (i)]
and b = ck
L.H.S. = ${\displaystyle \frac{a}{c}}$
= ${\displaystyle \frac{c{k}^2}{c}}$
= k²
R.H.S. = ${\displaystyle \frac{p{\left(c{k}^2\right)}^2+q\left(c{k}^2\right)ck+r{\left(ck\right)}^2}{p{\left(ck\right)}^2+q\left(ck\right)c+r{c}^2}}$

= ${\displaystyle \frac{p{c}^2{k}^4+q{c}^2{k}^3+r{c}^2{k}^2}{p{c}^2{k}^2+q{c}^{2}k+r{c}^2}}$

= ${\displaystyle \frac{c^2{k}^2}{c^2}\left[\frac{p{k}^2+qk+r}{p{k}^2+qk+r}\right]}$
= k²                              ...(iii)
From (ii) and (iii), L.H.s. = R.H.S.
∴ b = ck, a = bk = c k k = ck² 
(i) L.H.S.
= ${\displaystyle \frac{\text{a + b}}{\text{b + c}}}$

= ${\displaystyle \frac{c{k}^2+ck}{\text{ck + c}}}$

= ${\displaystyle \frac{ck(k+1)}{c(k+1)}}$
= k
R.H.S.
= ${\displaystyle \frac{a^2(b-c)}{b^2(a-b)}}$

= ${\displaystyle \frac{{\left(c{k}^2\right)}^2(ck-c)}{{\left(ck\right)}^2(c{k}^2-ck)}}$

= ${\displaystyle \frac{c^2{k}^{4}c(k-1)}{c^2{k}^2(k-1)}}$.