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 an arithmetic progression (AP) where the first three terms are given as \(3^{2a-1}\), \(14\), and \(3^{4-2a}\). We know that for three terms \(A\), \(B\), and \(C\) to be in AP, the condition \(2B = A + C\) must hold.

### Step-by-Step Solution:

1. **Set up the equation for AP**:
Given the terms:
\[ A = 3^{2a-1}, \quad B = 14, \quad C = 3^{4-2a} \]
The condition for these terms to be in AP is:
\[ 2B = A + C \]
Substituting the values:
\[ 2 \times 14 = 3^{2a-1} + 3^{4-2a} \]
This simplifies to:
\[ 28 = 3^{2a-1} + 3^{4-2a} \]

2. **Rewrite the terms using properties of exponents**:
Notice that \(3^{4-2a} = \frac{3^4}{3^{2a}} = \frac{81}{3^{2a}}\). Thus, we can rewrite the equation as:
\[ 28 = 3^{2a-1} + \frac{81}{3^{2a}} \]

3. **Let \(x = 3^{2a}\)**:
Then \(3^{2a-1} = \frac{x}{3}\) and the equation becomes:
\[ 28 = \frac{x}{3} + \frac{81}{x} \]

4. **Multiply through by \(3x\) to eliminate the fractions**:
\[ 28 \times 3x = x^2 + 243 \]
Simplifying gives:
\[ 84x = x^2 + 243 \]
Rearranging leads to:
\[ x^2 - 84x + 243 = 0 \]

5. **Solve the quadratic equation**:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{84 \pm \sqrt{(-84)^2 - 4 \cdot 1 \cdot 243}}{2 \cdot 1} \]
\[ x = \frac{84 \pm \sqrt{7056 - 972}}{2} \]
\[ x = \frac{84 \pm \sqrt{6084}}{2} \]
\[ x = \frac{84 \pm 78}{2} \]
This gives us two potential solutions:
\[ x = \frac{162}{2} = 81 \quad \text{and} \quad x = \frac{6}{2} = 3 \]

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

7. **Find the terms of the AP**:
Substitute \(a = \frac{1}{2}\):
\[ 3^{2a-1} = 3^{0} = 1, \quad 14, \quad 3^{4-2a} = 3^{3} = 27 \]
The first three terms are \(1, 14, 27\).

8. **Calculate the fifth term of the AP**:
The first term \(A = 1\) and the common difference \(D = 14 - 1 = 13\).
The \(n\)-th term of an AP is given by:
\[ T_n = A + (n-1)D \]
For the fifth term:
\[ T_5 = 1 + 4 \times 13 = 1 + 52 = 53 \]

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