IF `5^(th)` term of a H.P . Is `1/45` and `11^(th)` term is `1/69`, then its `16^(th)` term will be

Options

A. `1/89`
B. `1/85`
C. `1/80`
D. `1/79`

Answer

To find the 16th term of the harmonic progression (H.P.), we start by recalling that the terms of a harmonic progression can be expressed in terms of an arithmetic progression (A.P.). Specifically, if the terms of the H.P. are given by \( A_n \), then the corresponding terms of the A.P. can be represented as \( B_n = \frac{1}{A_n} \).

### Step 1: Define the terms
Let the first term of the A.P. be \( a \) and the common difference be \( d \). Then, the terms of the A.P. can be expressed as:
- \( B_5 = a + 4d \)
- \( B_{11} = a + 10d \)

### Step 2: Set up the equations
From the problem, we know:
- \( B_5 = \frac{1}{45} \) (5th term of H.P.)
- \( B_{11} = \frac{1}{69} \) (11th term of H.P.)

Thus, we can write the equations:
1. \( a + 4d = \frac{1}{45} \) (Equation 1)
2. \( a + 10d = \frac{1}{69} \) (Equation 2)

### Step 3: Solve the equations
To find \( a \) and \( d \), we can subtract Equation 1 from Equation 2:
\[ (a + 10d) - (a + 4d) = \frac{1}{69} - \frac{1}{45} \]
This simplifies to:
\[ 6d = \frac{1}{69} - \frac{1}{45} \]

### Step 4: Find a common denominator
The common denominator of 69 and 45 is 3105. Thus, we convert the fractions:
\[ \frac{1}{69} = \frac{45}{3105}, \quad \frac{1}{45} = \frac{69}{3105} \]
Now, substituting back:
\[ 6d = \frac{45 - 69}{3105} = \frac{-24}{3105} \]
So,
\[ d = \frac{-24}{6 \times 3105} = \frac{-4}{3105} \]

### Step 5: Substitute \( d \) back to find \( a \)
Now, substitute \( d \) back into Equation 1:
\[ a + 4\left(\frac{-4}{3105}\right) = \frac{1}{45} \]
This simplifies to:
\[ a - \frac{16}{3105} = \frac{1}{45} \]
Convert \( \frac{1}{45} \) to have a common denominator of 3105:
\[ \frac{1}{45} = \frac{69}{3105} \]
Thus:
\[ a - \frac{16}{3105} = \frac{69}{3105} \]
Adding \( \frac{16}{3105} \) to both sides gives:
\[ a = \frac{69 + 16}{3105} = \frac{85}{3105} \]

### Step 6: Find the 16th term
The 16th term of the H.P. corresponds to the 16th term of the A.P.:
\[ B_{16} = a + 15d \]
Substituting \( a \) and \( d \):
\[ B_{16} = \frac{85}{3105} + 15\left(\frac{-4}{3105}\right) \]
This simplifies to:
\[ B_{16} = \frac{85 - 60}{3105} = \frac{25}{3105} \]

### Step 7: Convert back to H.P.
The 16th term of the H.P. is:
\[ A_{16} = \frac{1}{B_{16}} = \frac{3105}{25} = 124.2 \]

### Final Answer:
The 16th term of the H.P. is \( \frac{3105}{25} \) or \( 124.2 \).