The expansion by the binomial theorem of `(2 x + (1)/(8) )^(10)` is `1024x^(10) + 640x^(9) + ax^(8) + bx^(7) + …` Calculate
(iii) the value of `x`, for which the third ·and the fourth terms in the expansion of `(2x + (1)/(8) )^(10)` are equal.

Answer

To solve the problem, we need to find the value of \( x \) for which the third and fourth terms in the expansion of \( (2x + \frac{1}{8})^{10} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = 2x \), \( b = \frac{1}{8} \), and \( n = 10 \).

2. **Write the Third and Fourth Terms**:
- The third term \( T_3 \) corresponds to \( r = 2 \):
\[ T_3 = \binom{10}{2} (2x)^{10-2} \left(\frac{1}{8}\right)^2 \]
- The fourth term \( T_4 \) corresponds to \( r = 3 \):
\[ T_4 = \binom{10}{3} (2x)^{10-3} \left(\frac{1}{8}\right)^3 \]

3. **Calculate \( T_3 \)**:
\[ T_3 = \binom{10}{2} (2x)^8 \left(\frac{1}{8}\right)^2 \]
\[ = \frac{10 \times 9}{2 \times 1} (2^8 x^8) \left(\frac{1}{64}\right) \]
\[ = 45 \cdot 256 x^8 \cdot \frac{1}{64} \]
\[ = 45 \cdot 4 x^8 = 180 x^8 \]

4. **Calculate \( T_4 \)**:
\[ T_4 = \binom{10}{3} (2x)^7 \left(\frac{1}{8}\right)^3 \]
\[ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} (2^7 x^7) \left(\frac{1}{512}\right) \]
\[ = 120 \cdot 128 x^7 \cdot \frac{1}{512} \]
\[ = 120 \cdot \frac{128}{512} x^7 = 120 \cdot \frac{1}{4} x^7 = 30 x^7 \]

5. **Set the Third and Fourth Terms Equal**:
\[ 180 x^8 = 30 x^7 \]

6. **Simplify the Equation**:
Divide both sides by \( 30 x^7 \) (assuming \( x \neq 0 \)):
\[ 6x = 1 \]

7. **Solve for \( x \)**:
\[ x = \frac{1}{6} \]

### Final Answer:
The value of \( x \) for which the third and fourth terms in the expansion of \( (2x + \frac{1}{8})^{10} \) are equal is:
\[ \boxed{\frac{1}{6}} \]