Question

In a H.P., if $5^{th}$ term is 6 and $3^{rd}$ term is 10. Find the $2^{nd}$ term.

• A. 15
• B. 12
• C. 14
• D. 16

Answer 1

Answer: (A)

15

Answer 2

To find the $2^{nd}$ term of a Harmonic Progression (H.P.), we first convert the problem into an Arithmetic Progression (A.P.) problem, as the reciprocals of terms in H.P. form an A.P.

Let the H.P. be $h_1, h_2, h_3, \dots$.
Then the corresponding A.P. is $a_1, a_2, a_3, \dots$, where $a_n = \frac{1}{h_n}$.

The $n^{th}$ term of an A.P. is given by the formula:
$a_n = A + (n-1)D$
where $A$ is the first term and $D$ is the common difference of the A.P.

Given information:

1. $5^{th}$ term of H.P. ($h_5$) = 6
   This implies the $5^{th}$ term of the corresponding A.P. ($a_5$) is $\frac{1}{6}$.
   So, $a_5 = A + (5-1)D = A + 4D = \frac{1}{6}$ (Equation 1)

2. $3^{rd}$ term of H.P. ($h_3$) = 10
   This implies the $3^{rd}$ term of the corresponding A.P. ($a_3$) is $\frac{1}{10}$.
   So, $a_3 = A + (3-1)D = A + 2D = \frac{1}{10}$ (Equation 2)

Now we have a system of two linear equations:

1. $A + 4D = \frac{1}{6}$
2. $A + 2D = \frac{1}{10}$

Subtract Equation 2 from Equation 1:
$(A + 4D) - (A + 2D) = \frac{1}{6} - \frac{1}{10}$
$2D = \frac{5 - 3}{30}$
$2D = \frac{2}{30}$
$2D = \frac{1}{15}$
$D = \frac{1}{30}$

Substitute the value of $D$ back into Equation 2:
$A + 2\left(\frac{1}{30}\right) = \frac{1}{10}$
$A + \frac{1}{15} = \frac{1}{10}$
$A = \frac{1}{10} - \frac{1}{15}$
$A = \frac{3 - 2}{30}$
$A = \frac{1}{30}$

So, the first term of the A.P. is $A = \frac{1}{30}$ and the common difference is $D = \frac{1}{30}$.

We need to find the $2^{nd}$ term of the H.P. ($h_2$). First, find the $2^{nd}$ term of the A.P. ($a_2$):
$a_2 = A + (2-1)D = A + D$
$a_2 = \frac{1}{30} + \frac{1}{30}$
$a_2 = \frac{2}{30}$
$a_2 = \frac{1}{15}$

Since $a_2 = \frac{1}{h_2}$, we have:
$\frac{1}{h_2} = \frac{1}{15}$
$h_2 = 15$