Question

If X,y,z are in continued proportion
then show that.
X-y upon x-z is equal to y upon y+z
​

Answer 1

Ans

\begin{gathered} \sf \frac{x - y}{x - z} = \frac{y}{y + z} \\ \end{gathered}

x−z

x−y

=

y+z

y

Given :-

X, Y, z are in continued proportion , it means that -

\sf \: x \ratio \: y \ratio \ratio \: y \ratio \:zx:y::y:z

SolutioN :-

\begin{gathered} \sf \: x \ratio \: y \ratio\ratio y\ratio \: z \\ \\ \sf \: we \: can \: write \: it \: as \mapsto\\ \\ \sf \implies \: \frac{x}{y} = \frac{y}{z} \\ \\ \sf \implies \: x \: \times \: z = y \times y \\ \\ \sf \implies \: xz = {y}^{2} \\ \\ \sf \implies \: {y}^{2} = xz - - - - (i)\end{gathered}

x:y::y:z

wecanwriteitas↦

⟹

y

x

=

z

y

⟹x×z=y×y

⟹xz=y

2

⟹y

2

=xz−−−−(i)

Now,

\begin{gathered} \sf \frac{x - y}{x - z} = \frac{y}{y + z} \\ \\ \implies \sf (x - y) \times (y + z) = y(x - z) \\ \\ \implies \sf \: xy + xz - {y}^{2} - yz = xy - yz \\ \\ \implies \sf \: \cancel{ xy} - \cancel{xy} + xz - \cancel{ yz }- {y}^{2} + \cancel{ yz} = 0 \\ \\ \implies \sf \: xz - {y}^{2} = 0\\ \\ \implies \sf \: {y}^{2} = xz \\ \\ \sf from \: equation \: (i) \mapsto \\ \\ \implies \sf \: xz = xz \\ \\ \implies \boxed{ \bold{lhs = rhs}}\end{gathered}

x−z

x−y

=

y+z

y

⟹(x−y)×(y+z)=y(x−z)

⟹xy+xz−y

2

−yz=xy−yz

⟹

xy

−

xy

+xz−

yz

−y

2

+

yz

=0

⟹xz−y

2

=0

⟹y

2

=xz

fromequation(i)↦

⟹xz=xz

⟹

lhs=rhs

______________________________

Hope it will help you :)

Answer 2

Answer:

O Bhai teri cutiepie anu to gai