Find the value of x, if 2+4+6+…+x=650.

Explanation

To find the value of \( x \) in the equation \( 2 + 4 + 6 + \ldots + x = 650 \), we can follow these steps:

### Step 1: Identify the series
The series given is an arithmetic series where:
- The first term \( a = 2 \)
- The common difference \( d = 4 - 2 = 2 \)

### Step 2: Determine the number of terms \( n \)
The last term of the series is \( x \). The \( n \)-th term of an arithmetic series can be expressed as:
\[ a_n = a + (n - 1) \cdot d \]
Setting \( a_n = x \), we have:
\[ x = 2 + (n - 1) \cdot 2 \]
This simplifies to:
\[ x = 2n \]

### Step 3: Use the formula for the sum of the first \( n \) terms
The sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
Substituting \( a = 2 \) and \( d = 2 \):
\[ S_n = \frac{n}{2} \cdot (2 \cdot 2 + (n - 1) \cdot 2) \]
This simplifies to:
\[ S_n = \frac{n}{2} \cdot (4 + 2n - 2) = \frac{n}{2} \cdot (2n + 2) = n(n + 1) \]

### Step 4: Set the sum equal to 650
We know from the problem statement that:
\[ n(n + 1) = 650 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ n^2 + n - 650 = 0 \]

### Step 6: Factor the quadratic equation
To solve the quadratic equation \( n^2 + n - 650 = 0 \), we can factor it:
\[ (n - 25)(n + 26) = 0 \]
This gives us two possible solutions:
\[ n - 25 = 0 \quad \Rightarrow \quad n = 25 \]
\[ n + 26 = 0 \quad \Rightarrow \quad n = -26 \quad (\text{not valid since } n \text{ must be positive}) \]

### Step 7: Calculate \( x \)
Now that we have \( n = 25 \), we can find \( x \):
\[ x = 2n = 2 \cdot 25 = 50 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{50} \]