There is a question in our book whose answer i am not being able to find out.: Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
This is what I have done so far:
Let the two numbers be $x$ and $y$. According to the question, $$\frac x{28}=\frac{28}y$$ $$\implies xy=28^2=784\tag1$$ And $$\frac xy=\frac y{224}$$ $$y^2=224x\tag2$$
What to do next?
let the two numbers are a and b and their mean proportion is c . let the third proportion of a and b is d . given c = 28 and d = 224 So a/28 = 28/b a*b = 28*28 --------------1) similarly a/b = b/224 so b^2 = 224a a = b^2 /224-----------2) from eq 1 and 2 b^3 = 28*28*224 = 7^3 *8^3 b = 56 so a = 28*28/56 = 14