If the sum of the series 2, 5, 8, 11,… is 60100, then n is

Options

A. 100
B. 200
C. 150
D. 250

Answer

To solve the problem, we need to find the number of terms \( n \) in the arithmetic series 2, 5, 8, 11, ... such that the sum of the series equals 60100.

### Step-by-step Solution:

1. **Identify the first term and common difference:**
- The first term \( a = 2 \).
- The common difference \( d = 5 - 2 = 3 \).

2. **Use the formula for the sum of the first \( n \) terms of an arithmetic series:**
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \times (2 \times 2 + (n - 1) \times 3) \]
Simplifying this:
\[ S_n = \frac{n}{2} \times (4 + 3n - 3) = \frac{n}{2} \times (3n + 1) \]

3. **Set the sum equal to 60100:**
\[ \frac{n}{2} \times (3n + 1) = 60100 \]
Multiplying both sides by 2 to eliminate the fraction:
\[ n(3n + 1) = 120200 \]

4. **Rearranging the equation:**
\[ 3n^2 + n - 120200 = 0 \]

5. **Use the quadratic formula to solve for \( n \):**
The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = 1 \), and \( c = -120200 \).
First, calculate the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \times 3 \times (-120200) = 1 + 1442400 = 1442401 \]
Now, calculate \( n \):
\[ n = \frac{-1 \pm \sqrt{1442401}}{2 \times 3} \]
Calculate \( \sqrt{1442401} \):
\[ \sqrt{1442401} = 1201 \]
Therefore:
\[ n = \frac{-1 \pm 1201}{6} \]

6. **Finding the two possible values for \( n \):**
- For the positive root:
\[ n = \frac{1200}{6} = 200 \]
- For the negative root:
\[ n = \frac{-1202}{6} \text{ (not valid since } n \text{ must be positive)} \]

7. **Conclusion:**
The number of terms \( n \) is \( 200 \).

### Final Answer:
\[ n = 200 \]