Question : Three numbers are such that their sum is 50, the product is 3750 and the sum of their reciprocals is $\frac{31}{150}$. Find the sum of the squares of the three numbers.
Option 1: $2500$
Option 2: $1250$
Option 3: $950$
Option 4: $122$
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Correct Answer: $950$
Solution :
Let the numbers be $a, b$ and $c$.
Given: $(a + b + c) = 50$, $abc = 3750$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{31}{150}$
Now, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{(bc + ac + ab)}{abc}$
⇒ $\text{(bc+ac +ab)} = \frac{31}{150}\times 3750 = 775$
Also, $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)$
⇒ $(50)^{2} = (a^{2} + b^{2} + c^{2}) + 2(775)$
⇒ $(a^{2} + b^{2} + c^{2}) = 2500 – 1550$
⇒ $(a^{2} + b^{2} + c^{2}) = 950$
Hence, the correct answer is $950$.
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