Answer :
To find the sum of the first 10 terms of an arithmetic sequence, we can follow these steps:
1. Identify the first term ([tex]\(a_1\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
- The first term ([tex]\(a_1\)[/tex]) is [tex]\(40\)[/tex].
- The sequence decreases by [tex]\(3\)[/tex] each time, so the common difference ([tex]\(d\)[/tex]) is [tex]\(-3\)[/tex].
2. Identify the number of terms ([tex]\(n\)[/tex]):
- We need to find the sum of the first [tex]\(10\)[/tex] terms, so [tex]\(n = 10\)[/tex].
3. Use the formula for the sum of the first [tex]\(n\)[/tex] terms of an arithmetic sequence:
[tex]\[ S_n = \frac{n}{2} \times [2a_1 + (n-1)d] \][/tex]
4. Plug in the identified values into the formula:
- First term, [tex]\(a_1 = 40\)[/tex]
- Common difference, [tex]\(d = -3\)[/tex]
- Number of terms, [tex]\(n = 10\)[/tex]
[tex]\[ S_{10} = \frac{10}{2} \times [2 \times 40 + (10-1) \times (-3)] \][/tex]
5. Simplify inside the brackets:
- Calculate [tex]\(2 \times 40\)[/tex]:
[tex]\[ 2 \times 40 = 80 \][/tex]
- Calculate [tex]\((10 - 1) \times (-3)\)[/tex]:
[tex]\[ 9 \times (-3) = -27 \][/tex]
- Add the results:
[tex]\[ 80 + (-27) = 53 \][/tex]
6. Now, plug back into the final sum formula:
[tex]\[ S_{10} = \frac{10}{2} \times 53 \][/tex]
- Simplify [tex]\(\frac{10}{2}\)[/tex]:
[tex]\[ \frac{10}{2} = 5 \][/tex]
- Multiply:
[tex]\[ 5 \times 53 = 265 \][/tex]
Therefore, the sum of the first 10 terms of the arithmetic sequence [tex]\(40, 37, 34, 31, \ldots\)[/tex] is [tex]\(265\)[/tex].
1. Identify the first term ([tex]\(a_1\)[/tex]) and the common difference ([tex]\(d\)[/tex]):
- The first term ([tex]\(a_1\)[/tex]) is [tex]\(40\)[/tex].
- The sequence decreases by [tex]\(3\)[/tex] each time, so the common difference ([tex]\(d\)[/tex]) is [tex]\(-3\)[/tex].
2. Identify the number of terms ([tex]\(n\)[/tex]):
- We need to find the sum of the first [tex]\(10\)[/tex] terms, so [tex]\(n = 10\)[/tex].
3. Use the formula for the sum of the first [tex]\(n\)[/tex] terms of an arithmetic sequence:
[tex]\[ S_n = \frac{n}{2} \times [2a_1 + (n-1)d] \][/tex]
4. Plug in the identified values into the formula:
- First term, [tex]\(a_1 = 40\)[/tex]
- Common difference, [tex]\(d = -3\)[/tex]
- Number of terms, [tex]\(n = 10\)[/tex]
[tex]\[ S_{10} = \frac{10}{2} \times [2 \times 40 + (10-1) \times (-3)] \][/tex]
5. Simplify inside the brackets:
- Calculate [tex]\(2 \times 40\)[/tex]:
[tex]\[ 2 \times 40 = 80 \][/tex]
- Calculate [tex]\((10 - 1) \times (-3)\)[/tex]:
[tex]\[ 9 \times (-3) = -27 \][/tex]
- Add the results:
[tex]\[ 80 + (-27) = 53 \][/tex]
6. Now, plug back into the final sum formula:
[tex]\[ S_{10} = \frac{10}{2} \times 53 \][/tex]
- Simplify [tex]\(\frac{10}{2}\)[/tex]:
[tex]\[ \frac{10}{2} = 5 \][/tex]
- Multiply:
[tex]\[ 5 \times 53 = 265 \][/tex]
Therefore, the sum of the first 10 terms of the arithmetic sequence [tex]\(40, 37, 34, 31, \ldots\)[/tex] is [tex]\(265\)[/tex].