To find the sum of all positive integers from 10 to 2,500 that are divisible by 10, we can approach this problem by identifying it as an arithmetic sequence where:

• The first term ($a$) is 10.
• The last term ($l$) is 2,500.
• The common difference ($d$) is 10.

Step 1: Identify the Sequence

The sequence of numbers divisible by 10 between 10 and 2,500 is: $10, 20, 30, \ldots, 2500$

Step 2: Find the Number of Terms

We use the formula for the $n$-th term of an arithmetic sequence:

$l = a + (n-1) \times d$

Plugging in the known values:

$2500 = 10 + (n-1) \times 10$

Simplify:

$2500 - 10 = (n-1) \times 10$

$2490 = (n-1) \times 10$

$n-1 = 249$

$n = 250$

So, there are 250 terms in the sequence.

Step 3: Use the Sum Formula for an Arithmetic Sequence

The sum $S$ of an arithmetic sequence is given by:

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

Substituting the values:

$S = \frac{250}{2} \times (10 + 2500)$

$S = 125 \times 2510$

$S = 313,750$

Thus, the sum of all positive integers from 10 to 2,500 that are divisible by 10 is 313,750.

Would you like further details or have any questions?

Related Questions:

1. What is the sum of all positive integers from 1 to 1,000 inclusive, which are divisible by 5?
2. How many positive integers between 100 and 1,000 are divisible by 20?
3. Find the sum of all positive even integers between 50 and 1,000.
4. How many terms are in the arithmetic sequence from 5 to 505 with a common difference of 5?
5. What is the sum of multiples of 15 between 15 and 300?

Tip:

When solving problems involving sums of sequences, identifying the first term, last term, and common difference is crucial for applying the correct formulas efficiently.