$\dfrac{a}{b}=\dfrac58+\dfrac bc=\dfrac{16}{25}$

$6 : x : : 12 : 36$

$x : 32 : : 32 : 64 $

$5 : 120 : : 40 : x$

Given,

Given, $4:5:: x:45$

$16 : 24 = x : 54$

Since they divide Rs. $1200$ in the ratio $2:3:5$ then let shares of them will be $2x,3x$ and $5x$.

In $ a:b::c:d;  d $ is the fourth proportional.

In, $ 14:21 :: 4 :d $, product of extremes $ = $ product of means

$ 14 \times d = 21 \times 4 $
$ d = 6 $

$a : b = b : 343$

$\dfrac{3}{27} = \dfrac{27}{b}$

$8 : x :: 16 : 35$

Given, the first three terms of a proportion are $3,5$ and $21$ respectively.

Let the fourth number be $'x'$,

So, Numbers proportion are $3, 5, 21, x$

According to proportionality,

Product of means $=$ Product of extremes

$3 : 5 :: 21 : x$

$\dfrac{3}{5} = \dfrac{21}{x}$

Cross multiply,

$3 \times x = 21 \times 5$

$3x = 105$

$x = 35$

Therefore, The required fourth number is $35$

$12 : 21 = 8 : x$

In $ a:b::c:d;   d $ is the fourth proportional.

In, $ 5: \sqrt {75} :: \sqrt {48} :d $, product of extremes $ = $ product of means

$  5 \times d = \sqrt {75} \times  \sqrt {48} $
$ d = \dfrac{\sqrt {75} \times  \sqrt {48}}{5} = \dfrac{5\sqrt {3} \times 4\sqrt {3}}{5} = 4 \times 3 = 12 $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 5:7 :: 8 :d $, product of extremes $ = $ product of means
$ 5 \times d = 7 \times 8 $
$ d = \dfrac {56}{5} = 11.2 $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 1.2:3.8 :: 9 :d $, product of extremes $ = $ product of mean
$ 1.2 \times d = 3.8 \times 9 $

$ d = 28.5 $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 2\dfrac{1}{2}:1 \dfrac{1}{4} :: 3\dfrac {1}{3} :d $, product of extremes $ = $ product of means
$ 2\dfrac{1}{2} \times d = 1 \dfrac{1}{4} \times 3\dfrac {1}{3} $ 

$ \dfrac{5}{2} \times d = \dfrac{5}{4} \times \dfrac {10}{3} $

$ d = \dfrac {\dfrac{5}{4} \times \dfrac {10}{3}}{ \dfrac{5}{2}} = \dfrac {5}{3} = 1 \dfrac {2}{3} $

If a : b :: b : c, then we say that a, b, c are in continued proportion, and
c is the third proportional of a and b.
Here, $ {b}^{2} = ac $ or $ c = \dfrac{{b}^{2}}{a} $
So, for $ 12, 16 $, the third proportional is $ c = \dfrac{{b}^{2}}{a} = \dfrac
{{16}^{2}}{16} = \dfrac{16 \times 16}{12} = \dfrac{64}{3} = 21\dfrac{1}{3} $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 3:42 :: 7 :d $, product of extremes $ = $ product of means
$ 3 \times d = 42 \times 7 $
$ d = 98 $

If a : b :: b : c, then we say that a, b, c are in continued proportion, and c is the third proportional of a and b.

Here, $ {b}^{2} = ac $ or $ b = \sqrt {ac} $

So, for $ 2, 8 $, the third proportional is $ b = \sqrt {ac} = \sqrt {2 \times 8} = \sqrt {16} = 4 $

If a : b :: b : c, then we say that a, b, c are in continued

proportion, and c is the third proportional of a and b.





Here, $ {b}^{2} = ac $ or $ b = \sqrt {ac} $





So, for $ 12, 192 $, the third proportional is $ b = \sqrt {12 \times 192} = \sqrt {4

\times 3 \times 64 \times 3} = 2 \times 3 \times 8 = 48 $

In $ a:b::c:d;   d $ is the fourth proportional.





For, $ 5:6 :: 7 :d $, product of extremes $ = $ product of means





$ 5 \times d = 6 \times 7 $


$ d = \frac {42}{5} = 8.40 $

Let third proportional is x
$\left ( x^{2}-y^{2} \right ):\left ( x-y \right )=\left ( x-y \right ):x$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left ( x^{2}-y^{2} \right )}$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left (x+y  \right )\left (x-y  \right )}$
$\Rightarrow x=\frac{\left ( x-y \right )}{\left (x+y  \right )}$
 

Let the third proportional to $\displaystyle (x^{2}-y^{2}): and: (x+y)$ be A.

To convert $\displaystyle \frac{1}{2}:\frac{1}{3}:\frac{1}{4}$ into normal ratio, multiply each fraction by the LCM$(2,3,4) = 12$

$\displaystyle= 12\times \frac{1}{2}:12\times\frac{1}{3}:12\times\frac{1}{4}$
= $\displaystyle 6:4:3 $
Thus, A would have got $\displaystyle \frac{6}{6+4+3} \times 117 = 54$

B would have got $\displaystyle \frac{4}{6+4+3} \times 117 = 36$

And C would have got $\displaystyle \frac{3}{6+4+3} \times 117 = 27$

Instead Rs. $117$ have been divided in the ratio of $2:3:4$

A gets $\displaystyle \frac {2}{2+3+4} \times 117 = 26$

B gets $\displaystyle \frac {3}{2+3+4} \times 117 = 39$

C gets $\displaystyle \frac {4}{2+3+4} \times 117 = 52$

Thus, by changing the ratio,
A gained $26 - 54$ = -$28$
B gained $39 - 36$ = $3$
And C gained $52 - 27 = 25$
Thus, C's gain of $25$ is the most.

Take the parts as x, y & z.

Let the parts be $7x,5x,3x$ and $4x.$ 

If a : b :: b : c, then we say that a, b, c are in continued

proportion, and c is the third proportional of a and b.





Here, $ {b}^{2} = ac $ or $ c = \frac{{b}^{2}}{a} $





So, for $ 8, 10 $, the third proportional is $ c = \frac{{b}^{2}}{a} = \frac {{10}^{2}}{8} = 12.50 $

If a : b :: b : c, then we say that a, b, c are in continued proportion, and

c is the third proportional of a and b.





Here, $ {b}^{2} = ac $ 





Given  the third proportional of $ 6, x $ is 150 

$ => {x}^{2} = 6 \times 150 = 6 \times 6 \times 25 $

$ => x = \sqrt { 6 \times 6 \times 25 } $

$ x = 6 \times 5 = 30 $

Work done by the inlet in 1 hour $= \dfrac{1}{6}.\dfrac{1}{4}=\dfrac{1}{24}$

$2:3 : :6:x$
$\Longrightarrow \dfrac{2}{3}=\dfrac{6}{x}$

Let the fourth proportional to $5,\,8,\,15$ be x
$5:8::15:x$

Required mean proportion $=\begin{pmatrix}234\times104\end{pmatrix}^{\dfrac{1}{2}}=156$

Given:

(i) Let the fourth proportional to $4, 9, 12$ be $x$

Let the third proportional to $0.36$ and $0.48$ be x
$0.36:0.48::0.48:x$

$Let\quad the\quad common\quad multiple\quad of\quad 2\quad and\quad 3\quad be\quad x.then:$

Given, $x + \cfrac {1}{3} x + \cfrac {1}{6}x = 78$