Question

If three quantities are in continued proporation show that the ratio of the first to the third is the duplicate ratio of the first to the second

Answer 1

let x ,y ,z ,be the three quantities which are in continued proportion

⇒x/y =y/z ..........,[1]

⇒y^2 =xz ..........[2]

squaring [1] we get

[x/y]^2 = [y/z]^2 ⇒ y^2/z^2 =x^2/y^2

⇒xz/y^2 =x^2 /y^2

⇒x/z =x^2/y^2

x : z =x^2 ; y^2

ie,the ratio of the first to the third is the duplicate ratio of the first to the second.

Answer 2

Step-by-step explanation:

Given:If three quantities are in continued proportion.

To find: Show that the ratio of the first to the third is the duplicate ratio of the first to the second.

Solution:

Let the three quantities are l,m and n

ATQ

ratio of first to third

$\frac{l}{n}...eq1$

duplicate ratio of first to second

$\frac{ {l}^{2} }{ {m}^{2} } ...eq2$

Equate both

$\frac{l}{n} = \frac{ {l}^{2} }{ {m}^{2} } \\ \\ \frac{1}{n} = \frac{l}{ {m}^{2} } \\ \\ or \\ \\ l = \frac{ {m}^{2} }{n} ...eq3$

Now, if l,m and n are in continuous proportion

then

$\frac{l}{m} = \frac{m}{n}...eq4$

put the value of eq3 in LHS of eq4

$\frac{ \frac{ {m}^{2} }{n} }{m} = \frac{m}{n} \\ \\ \frac{ {m}^{ \cancel2} }{n} \times \frac{1}{ \cancel m} = \frac{m}{n} \\ \\ \frac{m}{n} = \frac{m}{n}$

hence proved.

Final answer: If three quantities are in continued proportion than it has been shown that the ratio of the first to the third is the duplicate ratio of the first to the second.

Hope it helps you.