To find the sum of integers between 120 and 365 that are divisible by 9, follow these steps:

Step 1: Identify the first and last terms

1. First term: The smallest integer divisible by 9 greater than or equal to 120 is $126$ (since $9 \times 14 = 126$).
2. Last term: The largest integer divisible by 9 less than or equal to 365 is $360$ (since $9 \times 40 = 360$).

So, we are working with an arithmetic sequence that starts at $a = 126$ and ends at $l = 360$ with a common difference $d = 9$.

Step 2: Find the number of terms in the sequence

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

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

Set $a_n = 360$ and solve for $n$:

$360 = 126 + (n - 1) \times 9$ $234 = (n - 1) \times 9$ $n - 1 = \frac{234}{9} = 26$ $n = 27$

So, there are $27$ terms in this sequence.

Step 3: Calculate the sum of the sequence

The sum $S_n$ of an arithmetic sequence can be calculated with the formula:

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

Substitute $n = 27$, $a = 126$, and $l = 360$:

$S_{27} = \frac{27}{2} \times (126 + 360)$ $S_{27} = \frac{27}{2} \times 486$ $S_{27} = 27 \times 243 = 6561$

Answer

The sum of integers between 120 and 365 that are divisible by 9 is 6561.

Would you like further details or have any questions?

Here are some related questions to deepen understanding:

1. What is the general approach to find the sum of numbers in a sequence with any common divisor?
2. How would the approach differ if you needed numbers divisible by both 9 and another number?
3. Can you generalize this process to find sums within any arbitrary range?
4. What adjustments are needed if the sequence had non-integer terms or a fractional divisor?
5. How does the sum formula change if the sequence is not arithmetic but geometric?

Tip: For arithmetic sequences, verify both the first and last terms meet the conditions to avoid extra calculations.