Question 149

Three numbers are such that their sum is 50, product is 3750 and the sum of their reciprocals is $$\frac{31}{150}$$. Find the sum of the squares of the three numbers.

Solution

Let the numbers be $$x,y,z$$

Given : $$(x+y+z)=50$$ , $$xyz=3750$$ and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{31}{150}$$

Now, $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}$$

=> $$(xy+yz+zx)=\frac{31}{150}\times3750=775$$

$$\therefore$$ $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$

=> $$(50)^2=(x^2+y^2+z^2)+2(775)$$

=> $$x^2+y^2+z^2=2500-1550=950$$

=> Ans - (C)

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: