Certainly! Let's go through each problem step by step.
### 1. Finding the first term of the arithmetic sequence
Given:
- The 16th term ([tex]\(a_{16}\)[/tex]) is 48
- The common difference ([tex]\(d\)[/tex]) is 3
We use the formula for the n-th term of an arithmetic sequence:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
For the 16th term:
[tex]\[ 48 = a_1 + (16-1) \cdot 3 \][/tex]
[tex]\[ 48 = a_1 + 15 \cdot 3 \][/tex]
[tex]\[ 48 = a_1 + 45 \][/tex]
Solving for [tex]\(a_1\)[/tex]:
[tex]\[ a_1 = 48 - 45 \][/tex]
[tex]\[ a_1 = 3 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
### 2. Finding the 10th term of the arithmetic sequence
Given:
- [tex]\(a_3 = 38\)[/tex]
- [tex]\(a_9 = -4\)[/tex]
We need to determine the common difference ([tex]\(d\)[/tex]) first.
Using the formula for the n-th term:
[tex]\[ a_9 = a_3 + (9-3) \cdot d \][/tex]
[tex]\[ -4 = 38 + 6 \cdot d \][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[ -4 - 38 = 6 \cdot d \][/tex]
[tex]\[ -42 = 6 \cdot d \][/tex]
[tex]\[ d = -42 / 6 \][/tex]
[tex]\[ d = -7 \][/tex]
Now, to find [tex]\(a_{10}\)[/tex]:
[tex]\[ a_{10} = a_9 + d \][/tex]
[tex]\[ a_{10} = -4 + (-7) \][/tex]
[tex]\[ a_{10} = -11 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
### 3. Inserting 3 arithmetic means between 11 and -13
Given:
- Start term: 11
- End term: -13
- Number of means: 3
First, we calculate the number of intervals, which is one more than the number of means:
[tex]\[ \text{Intervals} = 3 + 1 = 4 \][/tex]
The common difference ([tex]\(d\)[/tex]) is determined by the difference between the end term and the start term divided by the number of intervals:
[tex]\[ d = \frac{-13 - 11}{4} \][/tex]
[tex]\[ d = \frac{-24}{4} \][/tex]
[tex]\[ d = -6 \][/tex]
Now, insert the arithmetic means:
1. 1st mean:
[tex]\[ 11 + (-6) = 5 \][/tex]
2. 2nd mean:
[tex]\[ 11 + 2(-6) = 11 - 12 = -1 \][/tex]
3. 3rd mean:
[tex]\[ 11 + 3(-6) = 11 - 18 = -7 \][/tex]
So, the correct arithmetic means are:
[tex]\[ 5, -1, -7 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
