If the numbers `3^(2a-1), 14, 3^(4-2a) (0 lt a lt 1)` are the first three terms of an arithmetic progression, then its fifth term is equal to

Options

A. 33
B. 43
C. 53
D. 63

Answer

To solve the problem, we need to determine the fifth term of the arithmetic progression (AP) formed by the numbers \(3^{2a-1}\), \(14\), and \(3^{4-2a}\) given that \(0 < a < 1\).

### Step-by-Step Solution:

1. **Identify the Terms of the AP**:
The first three terms of the AP are:
- First term \(T_1 = 3^{2a-1}\)
- Second term \(T_2 = 14\)
- Third term \(T_3 = 3^{4-2a}\)

2. **Use the Property of AP**:
For three numbers to be in AP, the following condition must hold:
\[ 2T_2 = T_1 + T_3 \]
Substituting the values we have:
\[ 2 \cdot 14 = 3^{2a-1} + 3^{4-2a} \]
This simplifies to:
\[ 28 = 3^{2a-1} + 3^{4-2a} \]

3. **Rewrite the Exponential Terms**:
We can rewrite \(3^{4-2a}\) as:
\[ 3^{4-2a} = \frac{81}{3^{2a}} \]
Thus, we have:
\[ 28 = 3^{2a-1} + \frac{81}{3^{2a}} \]

4. **Let \(x = 3^{2a}\)**:
Now, substituting \(x\) for \(3^{2a}\):
\[ 28 = \frac{x}{3} + \frac{81}{x} \]
Multiplying through by \(3x\) to eliminate the fractions gives:
\[ 84x = x^2 + 243 \]

5. **Rearrange to Form a Quadratic Equation**:
Rearranging the equation:
\[ x^2 - 84x + 243 = 0 \]

6. **Solve the Quadratic Equation**:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -84\), and \(c = 243\):
\[ x = \frac{84 \pm \sqrt{84^2 - 4 \cdot 1 \cdot 243}}{2 \cdot 1} \]
Calculate the discriminant:
\[ 84^2 = 7056, \quad 4 \cdot 243 = 972 \quad \Rightarrow \quad 7056 - 972 = 6084 \]
Now, taking the square root:
\[ \sqrt{6084} = 78 \]
Thus, we have:
\[ x = \frac{84 \pm 78}{2} \]
This gives us two potential solutions:
\[ x = \frac{162}{2} = 81 \quad \text{or} \quad x = \frac{6}{2} = 3 \]

7. **Determine \(a\)**:
Since \(x = 3^{2a}\):
- If \(x = 81\), then \(3^{2a} = 3^4 \Rightarrow 2a = 4 \Rightarrow a = 2\) (not valid since \(0 < a < 1\)).
- If \(x = 3\), then \(3^{2a} = 3^1 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}\) (valid).

8. **Find the First Three Terms**:
Substitute \(a = \frac{1}{2}\):
- First term: \(T_1 = 3^{2(\frac{1}{2}) - 1} = 3^{0} = 1\)
- Second term: \(T_2 = 14\)
- Third term: \(T_3 = 3^{4 - 2(\frac{1}{2})} = 3^{3} = 27\)

9. **Calculate the Fifth Term**:
The fifth term of an AP can be calculated using the formula:
\[ T_n = T_1 + (n-1)d \]
where \(d\) is the common difference. First, find \(d\):
\[ d = T_2 - T_1 = 14 - 1 = 13 \]
Now, calculate the fifth term:
\[ T_5 = T_1 + 4d = 1 + 4 \cdot 13 = 1 + 52 = 53 \]

### Final Answer:
The fifth term of the arithmetic progression is \(53\).