If the seventh term of an A.P. is 1/9 and its ninth term is 1/7 ,find its `63^(rd)` term .

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the problem, we need to find the 63rd term of an Arithmetic Progression (A.P.) given that the 7th term is \( \frac{1}{9} \) and the 9th term is \( \frac{1}{7} \).

### Step-by-Step Solution:

1. **Identify the General Formula for Terms in A.P.**:
The nth term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Set Up the Equations**:
For the 7th term:
\[ a_7 = a + 6d = \frac{1}{9} \quad \text{(Equation 1)} \]
For the 9th term:
\[ a_9 = a + 8d = \frac{1}{7} \quad \text{(Equation 2)} \]

3. **Subtract the Two Equations**:
Subtract Equation 1 from Equation 2 to eliminate \( a \):
\[ (a + 8d) - (a + 6d) = \frac{1}{7} - \frac{1}{9} \]
This simplifies to:
\[ 2d = \frac{1}{7} - \frac{1}{9} \]

4. **Find a Common Denominator**:
The least common multiple of 7 and 9 is 63. Rewrite the fractions:
\[ \frac{1}{7} = \frac{9}{63}, \quad \frac{1}{9} = \frac{7}{63} \]
Thus,
\[ 2d = \frac{9}{63} - \frac{7}{63} = \frac{2}{63} \]

5. **Solve for \( d \)**:
Divide both sides by 2:
\[ d = \frac{2}{63} \times \frac{1}{2} = \frac{1}{63} \]

6. **Substitute \( d \) Back to Find \( a \)**:
Substitute \( d \) back into Equation 1:
\[ a + 6 \left(\frac{1}{63}\right) = \frac{1}{9} \]
This simplifies to:
\[ a + \frac{6}{63} = \frac{1}{9} \]
Since \( \frac{6}{63} = \frac{2}{21} \), we rewrite:
\[ a + \frac{2}{21} = \frac{1}{9} \]

7. **Find a Common Denominator Again**:
The least common multiple of 9 and 21 is 63. Rewrite the fractions:
\[ \frac{1}{9} = \frac{7}{63}, \quad \frac{2}{21} = \frac{6}{63} \]
Thus,
\[ a + \frac{6}{63} = \frac{7}{63} \]

8. **Solve for \( a \)**:
Subtract \( \frac{6}{63} \) from both sides:
\[ a = \frac{7}{63} - \frac{6}{63} = \frac{1}{63} \]

9. **Find the 63rd Term**:
Now we can find the 63rd term using the formula:
\[ a_{63} = a + (63-1)d = a + 62d \]
Substitute \( a \) and \( d \):
\[ a_{63} = \frac{1}{63} + 62 \left(\frac{1}{63}\right) = \frac{1}{63} + \frac{62}{63} = \frac{63}{63} = 1 \]

### Final Answer:
The 63rd term of the A.P. is \( 1 \).