The given series is an arithmetic series where the first term is 1, the common difference is 5, and the last term is 100. To find the sum of an arithmetic series, we use the formula:

$S_n = \frac{n}{2} \times (a + l)$

Where:

• $S_n$ is the sum of the series,
• $n$ is the number of terms,
• $a$ is the first term,
• $l$ is the last term.

Step 1: Find the number of terms ($n$)

To find $n$, use the formula for the $n$-th term of an arithmetic sequence:

$a_n = a + (n-1) \cdot d$

Where:

• $a_n = 100$ (the last term),
• $a = 1$ (the first term),
• $d = 5$ (the common difference).

Solving for $n$:

$100 = 1 + (n-1) \cdot 5$ $100 - 1 = (n-1) \cdot 5$ $99 = (n-1) \cdot 5$ $n-1 = \frac{99}{5} = 19.8$

So $n = 21$.

Step 2: Calculate the sum

Now, substitute $n = 21$, $a = 1$, and $l = 100$ into the sum formula:

$S_{21} = \frac{21}{2} \times (1 + 100) = \frac{21}{2} \times 101 = 1050.5$

Hence, the sum is approximately 1050.5.

Would you like more details on this process, or help with related questions?

Related questions:

1. How do you find the sum of the first 100 terms of any arithmetic series?
2. How does the common difference affect the sum of an arithmetic series?
3. Can you apply the same method to a geometric series?
4. How do you derive the formula for the sum of an arithmetic sequence?
5. What happens if the common difference is negative in an arithmetic series?

Tip:

For large arithmetic series, knowing both the number of terms and the last term simplifies the calculation of the sum!