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

Answer

To solve the equation \(1 + 5 + 9 + 13 + \ldots + x = 1326\), we can follow these steps:

### Step 1: Identify the series
The series given is \(1, 5, 9, 13, \ldots\). We can see that this is an arithmetic progression (AP) where:
- The first term \(a = 1\)
- The common difference \(d = 5 - 1 = 4\)

### Step 2: Write the formula for the sum of an AP
The sum \(S_n\) of the first \(n\) terms of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
We know that \(S_n = 1326\).

### Step 3: Substitute the known values into the formula
Substituting \(a = 1\) and \(d = 4\) into the formula gives:
\[ 1326 = \frac{n}{2} \times (2 \times 1 + (n - 1) \times 4) \]
This simplifies to:
\[ 1326 = \frac{n}{2} \times (2 + 4n - 4) \]
\[ 1326 = \frac{n}{2} \times (4n - 2) \]
\[ 1326 = \frac{n(4n - 2)}{2} \]
\[ 1326 = 2n(2n - 1) \]

### Step 4: Rearrange the equation
Multiplying both sides by 2 to eliminate the fraction:
\[ 2652 = 2n(2n - 1) \]
This can be rearranged to:
\[ 4n^2 - 2n - 2652 = 0 \]

### Step 5: Solve the quadratic equation
Now we will solve the quadratic equation \(4n^2 - 2n - 2652 = 0\) using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 4\), \(b = -2\), and \(c = -2652\).

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4 \times 4 \times (-2652) = 4 + 42432 = 42436 \]
Now, calculate \(n\):
\[ n = \frac{2 \pm \sqrt{42436}}{8} \]
Calculating \(\sqrt{42436}\):
\[ \sqrt{42436} = 206 \]
So,
\[ n = \frac{2 \pm 206}{8} \]
Calculating the two possible values for \(n\):
1. \(n = \frac{208}{8} = 26\)
2. \(n = \frac{-204}{8}\) (not valid since \(n\) must be positive)

Thus, \(n = 26\).

### Step 6: Find the last term \(x\)
Now, we can find the last term \(x\) using the formula for the \(n\)th term of an AP:
\[ x = a + (n - 1)d \]
Substituting the values:
\[ x = 1 + (26 - 1) \times 4 \]
\[ x = 1 + 25 \times 4 \]
\[ x = 1 + 100 = 101 \]

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

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