Question

Three numbers are such that their sum is 50, product is 3750 and the sum of their reciprocals is 31 150 31150. Find the sum of the squares of the three numbers

Answer 1

Answer:

just try ur best

Step-by-step explanation:

Answer 2

Sum of the squares of the three numbers is 950

Solution:

Let the numbers be x, y, z

Given that,

Their sum is 50

x + y + z = 50 ------ eqn 1

Product is 3750

xyz = 3750 ------ eqn 3

The sum of their reciprocals is 31/150

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{31}{150}$

Simplify the above

$\frac{yz + xz + xy}{xyz} = \frac{31}{150}\\\\yz + xz + xy = \frac{31}{150} \times xyz\\\\Substitute\ eqn\ 3\\\\yz + xz + xy = \frac{31}{150} \times 3750\\\\yz + xz + xy = 31 \times 25\\\\yz + xz + xy = 775 -------- eqn\ 4$

We know that,

$(x+y+z)^2 = x^2+y^2+z^2 + 2(xy + yz + zx)\\\\Substitute\ eqn\ 1\\\\50^2 = x^2+y^2+z^2 + 2(xy + yz + zx)\\\\Substitute\ eqn\ 4\\\\50^2 = x^2+y^2+z^2 +2(775)\\\\2500 = x^2+y^2+z^2+1550\\\\x^2+y^2+z^2 = 2500-1550\\\\x^2+y^2+z^2 = 950$

Thus, sum of the squares of the three numbers is 950

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