Question

The sum of three numbers in H.P. is 11 and the sum of their reciprocals is 1, then their product is equal to

• A. 30
• B. 48
• C. 24
• D. 36

Answer 1

Answer: (D)

36

Answer 2

To solve this problem, we use the properties of Harmonic Progression (H.P.) and Arithmetic Progression (A.P.).

Let the three numbers in H.P. be $x, y, z$. If $x, y, z$ are in H.P., then their reciprocals $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.

Let the three numbers in A.P. be $a-d, a, a+d$. So, the three numbers in H.P. are $\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}$.

We are given two conditions:

1. The sum of the three numbers in H.P. is 11. $\frac{1}{a-d} + \frac{1}{a} + \frac{1}{a+d} = 11$ (Equation 1)

2. The sum of their reciprocals is 1. The reciprocals are $a-d, a, a+d$. $(a-d) + a + (a+d) = 1$ (Equation 2)

From Equation 2, we can find the value of $a$: $3a = 1$ $a = \frac{1}{3}$

Now, substitute the value of $a$ into Equation 1: $\frac{1}{\frac{1}{3}-d} + \frac{1}{\frac{1}{3}} + \frac{1}{\frac{1}{3}+d} = 11$ $\frac{3}{1-3d} + 3 + \frac{3}{1+3d} = 11$

Subtract 3 from both sides: $\frac{3}{1-3d} + \frac{3}{1+3d} = 11 - 3$ $\frac{3}{1-3d} + \frac{3}{1+3d} = 8$

Factor out 3 from the left side: $3 \left( \frac{1}{1-3d} + \frac{1}{1+3d} \right) = 8$

Combine the fractions inside the parenthesis: $3 \left( \frac{(1+3d) + (1-3d)}{(1-3d)(1+3d)} \right) = 8$ $3 \left( \frac{2}{1^2 - (3d)^2} \right) = 8$ $3 \left( \frac{2}{1 - 9d^2} \right) = 8$ $\frac{6}{1 - 9d^2} = 8$

Now, solve for $d$: $6 = 8(1 - 9d^2)$ $6 = 8 - 72d^2$ $72d^2 = 8 - 6$ $72d^2 = 2$ $d^2 = \frac{2}{72}$ $d^2 = \frac{1}{36}$ $d = \pm \sqrt{\frac{1}{36}}$ $d = \pm \frac{1}{6}$

Now we find the three numbers in H.P. using $a = \frac{1}{3}$ and $d = \pm \frac{1}{6}$.

Case 1: $d = \frac{1}{6}$ The terms in A.P. are: $a-d = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$ $a = \frac{1}{3}$ $a+d = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ So, the A.P. terms are $\frac{1}{6}, \frac{1}{3}, \frac{1}{2}$. The H.P. terms (reciprocals) are $6, 3, 2$.

Case 2: $d = -\frac{1}{6}$ The terms in A.P. are: $a-d = \frac{1}{3} - (-\frac{1}{6}) = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ $a = \frac{1}{3}$ $a+d = \frac{1}{3} + (-\frac{1}{6}) = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$ So, the A.P. terms are $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$. The H.P. terms (reciprocals) are $2, 3, 6$.

In both cases, the three numbers in H.P. are $2, 3, 6$.

Finally, we need to find their product: Product $= 2 \times 3 \times 6 = 6 \times 6 = 36$.