Answered

III. Assessment

Choose the letter of the correct answer.

1. The 16th term of an arithmetic sequence is 48. If the common difference is 3, find the first term.
A. 3
B. 2
C. 1
D. -1

2. If [tex]a_3 = 38[/tex] and [tex]a_9 = -4[/tex], find [tex]a_{10}[/tex].
A. -7
B. -11
C. -15
D. -17

3. Insert 3 arithmetic means between 11 and -13.
A. [tex] -5, -2, 1[/tex]
B. [tex] 5, -1, -7[/tex]
C. [tex] 4, 0, -4[/tex]
D. [tex] 6, 0, -6[/tex]



Answer :

GinnyAnswer
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]

Other Questions