Question

If the $7^{th}$ term of a H.P. is 8 and the $8^{th}$ term is 7. Then find the $28^{th}$ term.

• A. 3
• B. 2
• C. 4
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

2

Answer 2

To find the $28^{th}$ 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 an H.P. form an A.P.

Let the given H.P. be $H_1, H_2, \dots, H_n, \dots$.
We are given:
$H_7 = 8$
$H_8 = 7$

Let the corresponding A.P. be $A_1, A_2, \dots, A_n, \dots$, where $A_n = \frac{1}{H_n}$.
So, we have:
$A_7 = \frac{1}{H_7} = \frac{1}{8}$
$A_8 = \frac{1}{H_8} = \frac{1}{7}$

Let 'a' be the first term and 'd' be the common difference of this A.P.
The formula for the $n^{th}$ term of an A.P. is $A_n = a + (n-1)d$.

For the $7^{th}$ term:
$A_7 = a + (7-1)d \Rightarrow a + 6d = \frac{1}{8}$ --- (1)

For the $8^{th}$ term:
$A_8 = a + (8-1)d \Rightarrow a + 7d = \frac{1}{7}$ --- (2)

Now, we solve these two linear equations for 'a' and 'd'.
Subtract equation (1) from equation (2):
$(a + 7d) - (a + 6d) = \frac{1}{7} - \frac{1}{8}$
$d = \frac{8 - 7}{56}$
$d = \frac{1}{56}$

Substitute the value of 'd' back into equation (1):
$a + 6 \left(\frac{1}{56}\right) = \frac{1}{8}$
$a + \frac{6}{56} = \frac{1}{8}$
$a + \frac{3}{28} = \frac{1}{8}$
$a = \frac{1}{8} - \frac{3}{28}$
To subtract, find a common denominator, which is 56:
$a = \frac{7}{56} - \frac{6}{56}$
$a = \frac{1}{56}$

So, for the A.P., the first term $a = \frac{1}{56}$ and the common difference $d = \frac{1}{56}$.

Now, we need to find the $28^{th}$ term of the A.P., $A_{28}$:
$A_{28} = a + (28-1)d$
$A_{28} = a + 27d$
Substitute the values of 'a' and 'd':
$A_{28} = \frac{1}{56} + 27 \left(\frac{1}{56}\right)$
$A_{28} = \frac{1 + 27}{56}$
$A_{28} = \frac{28}{56}$
$A_{28} = \frac{1}{2}$

Finally, to find the $28^{th}$ term of the H.P., $H_{28}$, we take the reciprocal of $A_{28}$:
$H_{28} = \frac{1}{A_{28}}$
$H_{28} = \frac{1}{\frac{1}{2}}$
$H_{28} = 2$