Three numbers are in HP. Their sum is 37 and the sum of

Solution

Let the three numbers in Harmonic Progression (HP) be a,b,c. Then their reciprocals 1/a,1/b,1/c are in Arithmetic Progression (AP). This implies 2/b=1/a+1/c. We are given: 1.     Sum of the numbers: a+b+c=37 2.     Sum of their reciprocals: 1/a+1/b+1/c=1/4 From the property of AP reciprocals, 1/a+1/c=2/b. Substitute this into the second given equation: 2/b+1/b=1/4 3/b=1/4 b=12 Now substitute the value of b into the first given equation: a+12+c=37 a+c=37−12 a+c=25 We also have the relation 2/b=1/a+1/c: 2/12=1/a+1/c 1/6=(a+c)/(ac) Substitute a+c=25: 1/6=25/(ac) ac=25×6 ac=150 Now we have a system of two equations with two variables a and c: a+c=25 ac=150 Consider a quadratic equation with roots a and c: x²−(a+c)x+ac=0 x²−25x+150=0 We can solve this quadratic equation by factoring: (x−10)(x−15)=0 The roots are x=10 and x=15. So, {a,c}={10,15}. The three numbers in HP are a,b,c. Using the values we found, the numbers are 10, 12, 15 or 15, 12, 10.