ratio and proportion numerical

[deb696]

if a,b,c are in continued proportion , prove that

(a+b+c)²/(a²+b²+c²) =(a+b+c)/(a-b+c)

i am unable to find out how to proceed ... if i apply the formula by rearranging its not working .....

helP ....

 

[Deleted member 4993]

if a,b,c are in continued proportion , prove that

(a+b+c)²/(a²+b²+c²) =(a+b+c)/(a-b+c)

i am unable to find out how to proceed ... if i apply the formula by rearranging its not working .....

helP ....

According to your textbook/class-notes:

What is the definition of continued proportion?

 

[deb696]

rep

When quantities are in continued proportion, all the ratios are equal. If
a:b = b:c = c:d = d:e

 

[Deleted member 4993]

When quantities are in continued proportion, all the ratios are equal. If
a:b = b:c = c:d = d:e

So in your case

b² = ac

Show some work using that information in:

(a + b+ c)(a - b +c) = ??

 

[Deleted member 4993]

if a,b,c are in continued proportion , prove that

(a+b+c)²/(a²+b²+c²) =(a+b+c)/(a-b+c)

i am unable to find out how to proceed ... if i apply the formula by rearranging its not working .....

helP ....

Enough time has elapsed - so I submit my solution....

a,b,c are in continued proportion → b² = ac then

(a + b + c)(a - b + c) = (a + c)² - b² = a² + c² + 2ac - b² = a² + c² + 2b² - b² = a² + c² + b²

(a + b + c)/(a² + b² + c² ) = 1/(a - b + c)

(a + b + c)²/(a² + b² + c² ) = (a + b + c)/(a - b + c)