If `1/2,1/alpha_1 , 1/alpha_2` , ….., `1/alpha_20 , 1/6` are in A.P. and `1,beta_1,beta_2` ….., `beta_20` , 6 are in AP. Then, the number of digits in value of `alpha_18beta_3` is ____

Answer

To solve the problem, we need to find the values of \( \alpha_{18} \) and \( \beta_{3} \) from the given arithmetic progressions (APs) and then calculate \( \alpha_{18} \beta_{3} \).

### Step 1: Analyze the first AP
The first AP is given as:
\[ \frac{1}{2}, \frac{1}{\alpha_1}, \frac{1}{\alpha_2}, \ldots, \frac{1}{\alpha_{20}}, \frac{1}{6} \]
This sequence has 22 terms (from \( \frac{1}{2} \) to \( \frac{1}{6} \)).

Let:
- First term \( a = \frac{1}{2} \)
- Last term \( a_n = \frac{1}{6} \)
- Number of terms \( n = 22 \)

Using the formula for the \( n \)-th term of an AP:
\[ a_n = a + (n-1)d \]
Substituting the values:
\[ \frac{1}{6} = \frac{1}{2} + (22 - 1)d \]
\[ \frac{1}{6} = \frac{1}{2} + 21d \]

### Step 2: Solve for \( d \)
Rearranging gives:
\[ 21d = \frac{1}{6} - \frac{1}{2} \]
Finding a common denominator (6):
\[ \frac{1}{2} = \frac{3}{6} \implies 21d = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3} \]
Thus,
\[ d = -\frac{1}{3 \times 21} = -\frac{1}{63} \]

### Step 3: Find \( \alpha_{18} \)
Using the formula for the \( k \)-th term:
\[ \frac{1}{\alpha_{18}} = a + (18 - 1)d \]
Substituting \( a \) and \( d \):
\[ \frac{1}{\alpha_{18}} = \frac{1}{2} + 17\left(-\frac{1}{63}\right) \]
Calculating:
\[ \frac{1}{\alpha_{18}} = \frac{1}{2} - \frac{17}{63} \]
Finding a common denominator (126):
\[ \frac{1}{2} = \frac{63}{126} \quad \text{and} \quad \frac{17}{63} = \frac{34}{126} \]
Thus,
\[ \frac{1}{\alpha_{18}} = \frac{63}{126} - \frac{34}{126} = \frac{29}{126} \]
So,
\[ \alpha_{18} = \frac{126}{29} \]

### Step 4: Analyze the second AP
The second AP is given as:
\[ 1, \beta_1, \beta_2, \ldots, \beta_{20}, 6 \]
This sequence also has 22 terms.

Let:
- First term \( a = 1 \)
- Last term \( a_n = 6 \)

Using the same formula:
\[ 6 = 1 + (22 - 1)d \]
Thus,
\[ 6 = 1 + 21d \]
Rearranging gives:
\[ 21d = 6 - 1 = 5 \implies d = \frac{5}{21} \]

### Step 5: Find \( \beta_{3} \)
Using the formula for the \( k \)-th term:
\[ \beta_{3} = a + (3 - 1)d \]
Substituting \( a \) and \( d \):
\[ \beta_{3} = 1 + 2\left(\frac{5}{21}\right) = 1 + \frac{10}{21} = \frac{21}{21} + \frac{10}{21} = \frac{31}{21} \]

### Step 6: Calculate \( \alpha_{18} \beta_{3} \)
Now we can find \( \alpha_{18} \beta_{3} \):
\[ \alpha_{18} \beta_{3} = \left(\frac{126}{29}\right) \left(\frac{31}{21}\right) \]
Calculating:
\[ \alpha_{18} \beta_{3} = \frac{126 \times 31}{29 \times 21} \]
Calculating the numerator:
\[ 126 \times 31 = 3906 \]
Calculating the denominator:
\[ 29 \times 21 = 609 \]
Thus,
\[ \alpha_{18} \beta_{3} = \frac{3906}{609} \]
Simplifying:
\[ \frac{3906 \div 3}{609 \div 3} = \frac{1302}{203} \]

### Step 7: Final Calculation
Calculating \( 1302 \div 203 \):
\[ 1302 \approx 6.42 \quad \text{(which is approximately 6)} \]
Thus, \( \alpha_{18} \beta_{3} \) is approximately 6.

### Step 8: Count the digits
The number of digits in \( 6 \) is \( 1 \).

### Final Answer
The number of digits in the value of \( \alpha_{18} \beta_{3} \) is **1**.

---