Question

If the $10^{th}$ term of a HP is 21 and $21^{st}$ term of the same HP is 10, then find the $210^{th}$ term.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

Answer 2

To find the $210^{th}$ term of a Harmonic Progression (HP), we first convert the problem into an Arithmetic Progression (AP) problem.

1. Relation between HP and AP:

A sequence $h_1, h_2, h_3, \dots$ is in HP if its reciprocals, $\frac{1}{h_1}, \frac{1}{h_2}, \frac{1}{h_3}, \dots$, form an Arithmetic Progression (AP).
Let the given HP be $h_n$ and the corresponding AP be $a_n$, such that $a_n = \frac{1}{h_n}$.
The general term of an AP is given by $a_n = A + (n-1)D$, where $A$ is the first term and $D$ is the common difference.

2. Formulate equations for the corresponding AP terms:

Given:
$10^{th}$ term of HP ($h_{10}$) is 21.
So, the $10^{th}$ term of the corresponding AP ($a_{10}$) is $\frac{1}{21}$.
Using the AP formula:
$a_{10} = A + (10-1)D = A + 9D = \frac{1}{21}$ (Equation 1)

$21^{st}$ term of HP ($h_{21}$) is 10.
So, the $21^{st}$ term of the corresponding AP ($a_{21}$) is $\frac{1}{10}$.
Using the AP formula:
$a_{21} = A + (21-1)D = A + 20D = \frac{1}{10}$ (Equation 2)

3. Solve the system of equations to find A and D:

Subtract Equation 1 from Equation 2:
$(A + 20D) - (A + 9D) = \frac{1}{10} - \frac{1}{21}$
$11D = \frac{21 - 10}{210}$
$11D = \frac{11}{210}$
$D = \frac{1}{210}$

Substitute the value of $D$ into Equation 1:
$A + 9\left(\frac{1}{210}\right) = \frac{1}{21}$
$A + \frac{9}{210} = \frac{1}{21}$
$A + \frac{3}{70} = \frac{1}{21}$
$A = \frac{1}{21} - \frac{3}{70}$
To subtract, find the Least Common Multiple (LCM) of 21 and 70, which is 210.
$A = \frac{1 \times 10}{21 \times 10} - \frac{3 \times 3}{70 \times 3}$
$A = \frac{10}{210} - \frac{9}{210}$
$A = \frac{1}{210}$

So, for the corresponding AP, the first term $A = \frac{1}{210}$ and the common difference $D = \frac{1}{210}$.

4. Calculate the required term of the AP:

We need to find the $210^{th}$ term of the HP, which means we first find the $210^{th}$ term of the corresponding AP ($a_{210}$).
$a_{210} = A + (210-1)D$
$a_{210} = A + 209D$
Substitute the values of A and D:
$a_{210} = \frac{1}{210} + 209\left(\frac{1}{210}\right)$
$a_{210} = \frac{1 + 209}{210}$
$a_{210} = \frac{210}{210}$
$a_{210} = 1$

5. Convert back to the HP term:

The $210^{th}$ term of the HP ($h_{210}$) is the reciprocal of the $210^{th}$ term of the AP ($a_{210}$).
$h_{210} = \frac{1}{a_{210}}$
$h_{210} = \frac{1}{1}$
$h_{210} = 1$

The $210^{th}$ term of the HP is 1.