Question

Suppose a is a fixed real number such that

$\frac{a - x}{px} = \frac{a -y }{qy} = \frac{a - z}{rz}$
If p, q, r are in AP, prove that x, y, z are in HP​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• p, q, r are in AP

$\rm \implies\: q - p = r - q$

Now, further given that

$\rm \: \dfrac{a - x}{px} = \dfrac{a -y }{qy} = \dfrac{a - z}{rz}$

can be further rewritten as

$\rm \: \dfrac{ \dfrac{a}{x} - 1}{p} = \dfrac{ \dfrac{a}{y} -1}{q} = \dfrac{ \dfrac{a}{z} - 1}{r} - - - (1)$

Consider,

$\rm \: \dfrac{ \dfrac{a}{x} - 1}{p} = \dfrac{ \dfrac{a}{y} -1}{q}$

can be rewritten as

$\rm \: \dfrac{ \dfrac{a}{x} - 1}{p} = \dfrac{1 - \dfrac{a}{y}}{ - q}$

Now, Using Addendo, we get

$\rm \: \dfrac{ \dfrac{a}{x} - 1}{p} = \dfrac{1 - \dfrac{a}{y}}{ - q} = \dfrac{\dfrac{a}{x} - 1 + 1 - \dfrac{a}{y} }{ p- q}$

$\rm\implies \:\rm \: \dfrac{ \dfrac{a}{x} - 1}{p} = \dfrac{\dfrac{a}{x} - \dfrac{a}{y} }{ p- q} - - - (2)$

Now, Consider

$\rm \: \dfrac{ \dfrac{a}{y} -1}{q} = \dfrac{ \dfrac{a}{z} - 1}{r}$

can be rewritten as

$\rm \: \dfrac{ \dfrac{a}{y} -1}{q} = \dfrac{1 - \dfrac{a}{z}}{ - r}$

On applying Addendo, we get

$\rm \: \dfrac{ \dfrac{a}{y} -1}{q} = \dfrac{1 - \dfrac{a}{z}}{ - r} = \dfrac{\dfrac{a}{y} - 1 + 1 - \dfrac{a}{z} }{q - r}$

$\rm\implies \:\rm \: \dfrac{ \dfrac{a}{y} -1}{q} = \dfrac{\dfrac{a}{y} - \dfrac{a}{z} }{q - r} - - - (3)$

Now, From equation (1), (2) and (3), we concluded that

$\rm \: \dfrac{ \dfrac{a}{x} - \dfrac{a}{y} }{p - q} = \dfrac{\dfrac{a}{y} - \dfrac{a}{z} }{q - r}$

can be further rewritten as

$\rm \: \dfrac{ \dfrac{a}{x} - \dfrac{a}{y} }{q - p} = \dfrac{\dfrac{a}{y} - \dfrac{a}{z} }{r - q}$

$\rm \: \dfrac{a}{x} - \dfrac{a}{y} = \dfrac{a}{y} - \dfrac{a}{z} \: \: \: \: \: \: \: \: \: \{ \because \: q - p = q - r \}$

$\rm \: \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{y} - \dfrac{1}{z}$

$\rm\implies \:\rm \: \dfrac{1}{x}, \: \dfrac{1}{y}, \: \dfrac{1}{z} \: are \: in \:AP$

$\rm\implies \:\rm \: x, \: y, \: z \: are \: in \:HP$

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Remark :-

Addendo :-

$\rm \: \dfrac{a}{b} = \dfrac{c}{d} \: \bf\implies \: \: \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a + b}{c + d}$

Additional Information :-

Alternendo :-

$\rm \: \dfrac{a}{b} = \dfrac{c}{d} \: \bf\implies \: \: \dfrac{a}{c} = \dfrac{b}{d}$

Invertendo :-

$\rm \: \dfrac{a}{b} = \dfrac{c}{d} \: \bf\implies \: \: \dfrac{b}{a} = \dfrac{d}{c}$

Componendo :-

$\rm \: \dfrac{a}{b} = \dfrac{c}{d} \: \bf\implies \: \: \dfrac{a + b}{b} = \dfrac{c + d}{d}$

Dividendo :-

$\rm \: \dfrac{a}{b} = \dfrac{c}{d} \: \bf\implies \: \: \dfrac{a - b}{b} = \dfrac{c - d}{d}$

Componendo and Dividendo :-

$\rm \: \dfrac{a}{b} = \dfrac{c}{d} \: \bf\implies \: \: \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

Answer 2

Answer:

Question :-

Suppose a is a fixed real number such that $\displaystyle \rm \frac{a - x}{px} = \frac{a -y }{qy} = \frac{a - z}{rz}$

If p, q, r are in AP, prove that x, y, z are in HP

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$\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

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$\text{\( \rm \because \quad p , q , r \) are in A.P.,}$

$\rm\therefore \quad q - p = r - q$

$\text{ \( \rm \Rightarrow \quad p - q = q - r = k \) (Let)}$

$\text{ Given, \( \displaystyle\rm \frac{a-x}{p x}=\frac{a-y}{q y}=\frac{a-z}{r z} \)}$

$\[ \begin{array}{l} \displaystyle\rm\Rightarrow \frac{\frac{ a }{ x }-1}{ p }=\frac{\frac{ a }{ y }-1}{ q }=\frac{\frac{ a }{ z }-1}{ r } \\ \\ \displaystyle\rm\Rightarrow \frac{\left(\dfrac{ a }{ x }-1\right)-\left(\dfrac{ a }{ y }-1\right)}{ p - q }=\frac{\left(\dfrac{ a }{ y }-1\right)-\left(\dfrac{ a }{ z }-1\right)}{ q - r } \\ \\ \text{ (by law of proportion)} \end{array} \]$

$\[ \begin{array}{l} \\ \displaystyle\rm \Rightarrow \frac{\frac{ a }{ x }-\frac{ a }{ y }}{ k }=\frac{\frac{ a }{ y }-\frac{ a }{ z }}{ k } \: \: \: \: \: \: \qquad \text { (from (i)) } \\\\ \displaystyle\rm \Rightarrow a \left(\frac{1}{ x }-\frac{1}{ y }\right)= a \left(\frac{1}{ y }-\frac{1}{ z }\right) \\\\ \displaystyle\rm \Rightarrow \frac{1}{ x }-\frac{1}{ y }=\frac{1}{ y }-\frac{1}{ z } \\ \\ \displaystyle\rm\therefore \frac{2}{ y }=\frac{1}{ x }+\frac{1}{ z } \\\\ \displaystyle\rm \therefore \frac{1}{ x }, \frac{1}{ y }, \frac{1}{ z } \text { are in A. P. } \end{array} \]$

Hence x, y, z are in H.P.