Math, asked by aryan021212, 19 days ago

If
 \frac{x}{y + z} \: and \:  \frac{y}{z + x}  \: and \:  \frac{z}{x + y}

are in AP, prove that x, y, z are in AO​

Answers

Answered by mathdude500
25

Appropriate Question

\rm \: If \: \dfrac{x}{y + z}, \:  \dfrac{y}{z + x}, \:  \dfrac{z}{x + y}  \: are \: in \: AP, \:  \\ \rm \:  \: prove \: that \:  {x}^{2}, {y}^{2}, {z}^{2} \: are \: in \: AP \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

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

Given that

\rm \: \dfrac{x}{y + z}, \:  \dfrac{y}{z + x}, \:  \dfrac{z}{x + y}  \: are \: in \: AP

We know, if each term is increased by any real number k, the resultant number are in AP

On adding 1 in each term, we get

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

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

On dividing each term by x + y + z, we get

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

We know, 3 number a, b, c are in AP, iff b - a = c - b

So,

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

\rm \: \dfrac{y + z - z - x}{(z + x)(y + z)}  = \dfrac{z + x - x - y}{(x + y)(z + x)}

\rm \: \dfrac{y - x}{y + z}  = \dfrac{z- y}{x + y}

\rm \: (y + x)(y - x) = (z + y)(z - y)

\rm \:  {y}^{2} -  {x}^{2} =  {z}^{2} -  {y}^{2}

\rm\implies \: {x}^{2}, \:  {y}^{2}, \:  {z}^{2}  \: are \: in \: AP

Hence, Proved

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ADDITIONAL INFORMATION

if each term is decreased by any real number k, the resultant number are in AP

if each term is multiplied by any real number k, the resultant number are in AP

if each term is divided by any real number k, the resultant number are in AP

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