Here is a problem involving the sum of the terms of an arithmetic sequence:

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Problem:

A gardener plants 10 trees in a straight line. The distance between the first two trees is 3 meters, and for each subsequent pair of trees, the distance increases by 1 meter.

• What is the total distance between the first tree and the last tree?

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Solution Outline:

1. Step 1: Identify the arithmetic sequence
   The distances between the trees form an arithmetic sequence:

   • First distance (between the 1st and 2nd tree): 3 meters
   • Common difference: 1 meter

2. Step 2: Number of terms
   The gardener plants 10 trees, so there are 9 spaces between them, making the number of terms in the sequence 9.

3. Step 3: Use the sum formula for an arithmetic sequence
   The sum $S_n$ of the first $n$ terms of an arithmetic sequence is given by: $S_n = \frac{n}{2} \cdot (2a + (n - 1) d)$ where:

   • $n$ is the number of terms,
   • $a$ is the first term,
   • $d$ is the common difference.

4. Step 4: Plug in the values
   In this problem:

   • $a = 3$ (first distance),
   • $d = 1$,
   • $n = 9$.

   So,

   = \frac{9}{2} \cdot (6 + 8) = \frac{9}{2} \cdot 14 = 63 \, \text{meters}.$$

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Answer: The total distance between the first tree and the last tree is 63 meters.

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Would you like more details on any part of this? Here are some related questions for further practice:

1. If the initial distance was 5 meters and increased by 2 meters, what would the total distance be?
2. How many trees would be needed to reach a total distance of 100 meters with the same arithmetic sequence?
3. If the gardener planted 15 trees with a starting distance of 2 meters and a common difference of 0.5 meters, what would the total distance be?
4. What is the distance between the first tree and the last tree if the gardener plants 12 trees with a starting distance of 4 meters and an increase of 1.5 meters?
5. How would the total distance change if the common difference was halved?

Tip: For arithmetic sequences, the sum formula is especially useful when dealing with long sequences, as it avoids adding each term individually.