Question

if a b c d are in continued proportion then(ma^3+nb^3-rc^3):(mb^3+nc^3-rd^3)=

Answer 1

Answer on Question # 41422 – Math - Other

if a b c d are in continued proportion then(ma^3+nb^3-rc^3):(mb^3+nc^3-rd^3)=

Solution.

If $a, b, c, d$ are in continued proportion, then

$$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}; \qquad \frac{a}{b} = \frac{c}{d} \Rightarrow \frac{d}{a} = \frac{bc}{a^2}, \quad \frac{d}{b} = \frac{c}{a}$$

Consider our fraction:

$$\frac{ma^3 + nb^3 - rc^3}{mb^3 + nc^3 - rd^3} = \frac{ma^3}{mb^3 + nc^3 - rd^3} + \frac{nb^3}{mb^3 + nc^3 - rd^3} - \frac{rc^3}{mb^3 + nc^3 - rd^3}$$

Consider the first term:

$$\frac{ma^3}{mb^3 + nc^3 - rd^3} = \frac{m}{m \frac{b^3}{a^3} + n \frac{c^3}{a^3} - r \frac{d^3}{a^3}} = \frac{m}{m \frac{b^3}{a^3} + n \frac{c^3}{a^3} - r \frac{b^3 c^3}{a^6}}$$

The second term:

$$\frac{nb^3}{mb^3 + nc^3 - rd^3} = \frac{n}{m + n \frac{c^3}{b^3} - r \frac{d^3}{b^3}} = \frac{n}{m + n \frac{b^3}{a^3} - r \frac{c^3}{a^3}}$$

The third term:

$$\frac{rc^3}{mb^3 + nc^3 - rd^3} = \frac{r}{m \frac{b^3}{c^3} + n - r \frac{d^3}{c^3}} = \frac{r}{m \frac{a^3}{b^3} + n - r \frac{b^3}{a^3}}$$

Let $\frac{b^3}{a^3} = x, \frac{c^3}{a^3} = y$. Then $\frac{y}{x} = \frac{c^3}{b^3} = \frac{b^3}{a^3} = x \Rightarrow y = x^2$:

$$\begin{array}{l} \frac{ma^3 + nb^3 - rc^3}{mb^3 + nc^3 - rd^3} = \frac{m}{mx + ny - rxy} + \frac{n}{m + nx - ry} + \frac{r}{\frac{m}{x} + n - rx} \\ = \frac{m}{mx + nx^2 - rx^3} + \frac{n}{m + nx - rx^2} + \frac{\frac{r}{m}}{x + n - rx} \\ = \frac{m}{mx + nx^2 - rx^3} + \frac{nx}{mx + nx^2 - rx^3} + \frac{rx^2}{mx + nx^2 - rx^3} = \frac{m + nx + rx^2}{x(m + nx - rx^2)} \\ = \frac{m + nx - rx^2 + 2rx^2}{x(m + nx - rx^2)} = \frac{1}{x} + \frac{2rx}{m + nx - rx^2} = \frac{1}{x} + \frac{2r}{\frac{m}{x} + n - rx} \end{array}$$

So

$$\frac{ma^3 + nb^3 - rc^3}{mb^3 + nc^3 - rd^3} = \frac{a^3}{b^3} + \frac{2r}{\left(m \frac{a^3}{b^3} + n - r \frac{b^3}{a^3}\right)}$$

Answer: $\frac{a^3}{b^3} + \frac{2r}{\left(m \frac{a^3}{b^3} + n - r \frac{b^3}{a^3}\right)}$

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