Solve the equation: 1+ 5+9+13+...+x= 1326.

Answer

To solve the equation \( 1 + 5 + 9 + 13 + \ldots + x = 1326 \), we first recognize that the series is an arithmetic progression (AP).

### Step-by-step Solution:

1. **Identify the first term and the common difference**:
- The first term \( a = 1 \).
- The common difference \( d = 5 - 1 = 4 \).

2. **Write the general formula for the nth term of an AP**:
- The nth term \( a_n \) of an AP can be expressed as:
\[ a_n = a + (n - 1)d \]
- Substituting the values of \( a \) and \( d \):
\[ a_n = 1 + (n - 1) \cdot 4 = 1 + 4n - 4 = 4n - 3 \]

3. **Sum of the first n terms of an AP**:
- The sum \( S_n \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1)d) \]
- Substituting the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \cdot (2 \cdot 1 + (n - 1) \cdot 4) = \frac{n}{2} \cdot (2 + 4n - 4) = \frac{n}{2} \cdot (4n - 2) \]
- Simplifying further:
\[ S_n = n(2n - 1) \]

4. **Set the sum equal to 1326**:
- We have:
\[ n(2n - 1) = 1326 \]

5. **Rearranging the equation**:
- Rearranging gives:
\[ 2n^2 - n - 1326 = 0 \]

6. **Using the quadratic formula**:
- The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 2, b = -1, c = -1326 \):
\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1326)}}{2 \cdot 2} \]
- This simplifies to:
\[ n = \frac{1 \pm \sqrt{1 + 10608}}{4} = \frac{1 \pm \sqrt{10609}}{4} \]
- Since \( \sqrt{10609} = 103 \):
\[ n = \frac{1 \pm 103}{4} \]

7. **Calculating the possible values for n**:
- This gives two potential solutions:
\[ n = \frac{104}{4} = 26 \quad \text{and} \quad n = \frac{-102}{4} = -25.5 \]
- Since \( n \) must be a natural number, we take \( n = 26 \).

8. **Finding the last term \( x \)**:
- Now, we find the last term \( a_n \) when \( n = 26 \):
\[ x = a_{26} = 1 + (26 - 1) \cdot 4 = 1 + 25 \cdot 4 = 1 + 100 = 101 \]

### Final Answer:
Thus, the value of \( x \) is \( 101 \).

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