Determine k so that:
(i) `k+2,4k-6,3k-2`
(ii) `8k+4,6k-2,2k-7`
(iii) `2/3,k, 5/8`
are the three consecutive terms of an A.P.

Answer

To determine the value of \( k \) such that the given sets of terms are in Arithmetic Progression (A.P.), we will apply the property of A.P. which states that for three terms \( a, b, c \) to be in A.P., the middle term \( b \) must be equal to the average of the other two terms, i.e., \( b = \frac{a + c}{2} \).

### Part (i): \( k + 2, 4k - 6, 3k - 2 \)

1. **Set up the equation using the A.P. property**:
\[ 4k - 6 = \frac{(k + 2) + (3k - 2)}{2} \]

2. **Simplify the right side**:
\[ 4k - 6 = \frac{(k + 3k + 2 - 2)}{2} = \frac{4k}{2} = 2k \]

3. **Set up the equation**:
\[ 4k - 6 = 2k \]

4. **Rearranging the equation**:
\[ 4k - 2k = 6 \implies 2k = 6 \]

5. **Solve for \( k \)**:
\[ k = 3 \]

### Part (ii): \( 8k + 4, 6k - 2, 2k - 7 \)

1. **Set up the equation using the A.P. property**:
\[ 6k - 2 = \frac{(8k + 4) + (2k - 7)}{2} \]

2. **Simplify the right side**:
\[ 6k - 2 = \frac{(8k + 2k + 4 - 7)}{2} = \frac{10k - 3}{2} \]

3. **Set up the equation**:
\[ 6k - 2 = \frac{10k - 3}{2} \]

4. **Multiply both sides by 2 to eliminate the fraction**:
\[ 2(6k - 2) = 10k - 3 \implies 12k - 4 = 10k - 3 \]

5. **Rearranging the equation**:
\[ 12k - 10k = -3 + 4 \implies 2k = 1 \]

6. **Solve for \( k \)**:
\[ k = \frac{1}{2} \]

### Part (iii): \( \frac{2}{3}, k, \frac{5}{8} \)

1. **Set up the equation using the A.P. property**:
\[ k = \frac{\frac{2}{3} + \frac{5}{8}}{2} \]

2. **Find a common denominator to add the fractions**:
The least common multiple of 3 and 8 is 24.
\[ \frac{2}{3} = \frac{16}{24}, \quad \frac{5}{8} = \frac{15}{24} \]

3. **Add the fractions**:
\[ k = \frac{\frac{16}{24} + \frac{15}{24}}{2} = \frac{\frac{31}{24}}{2} = \frac{31}{48} \]

### Final Values of \( k \):
- For part (i): \( k = 3 \)
- For part (ii): \( k = \frac{1}{2} \)
- For part (iii): \( k = \frac{31}{48} \)