Find the value of x for which 2x, (x + 10) and (3x + 2) are three consecutive terms of an A.P.

Answer

To find the value of \( x \) for which \( 2x \), \( x + 10 \), and \( 3x + 2 \) are three consecutive terms of an Arithmetic Progression (A.P.), we can use the property of A.P. that states the middle term is the average of the other two terms.

### Step-by-Step Solution:

1. **Identify the Terms**:
The three terms given are:
- First term: \( a = 2x \)
- Second term: \( b = x + 10 \)
- Third term: \( c = 3x + 2 \)

2. **Use the A.P. Property**:
For three terms to be in A.P., the following condition must hold:
\[ 2b = a + c \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ 2(x + 10) = 2x + (3x + 2) \]

3. **Expand Both Sides**:
Expanding the left side:
\[ 2x + 20 = 2x + 3x + 2 \]

4. **Combine Like Terms**:
On the right side, combine the terms:
\[ 2x + 20 = 5x + 2 \]

5. **Rearrange the Equation**:
To isolate \( x \), we can rearrange the equation:
\[ 20 - 2 = 5x - 2x \]
This simplifies to:
\[ 18 = 3x \]

6. **Solve for \( x \)**:
Divide both sides by 3:
\[ x = \frac{18}{3} = 6 \]

### Final Answer:
Thus, the value of \( x \) is \( 6 \).

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