Question
The first four terms of the sequence with \( a_{1}=40 \) and \( a_{n}=\frac{3}{4} a_{n-1} \) are A. \( 30,22,17,13 \) B. \( 40,30,22 \frac{1}{2}, 16 \frac{7}{8} \) C. \( 40,30,22,17 \) D. \( 30,22 \frac{1}{2}, 16 \frac{7}{8}, 12 \frac{21}{33} \) Option 1 Option 2 Option 3 Option 4
Ask by George Schultz. in the United States
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The first four terms are \( 40, 30, 22\frac{1}{2}, 16\frac{7}{8} \), which is Option B.
Solution
We are given:
\[
a_1 = 40\quad\text{and}\quad a_n = \frac{3}{4}a_{n-1}.
\]
1. The first term is:
\[
a_1 = 40.
\]
2. The second term is:
\[
a_2 = \frac{3}{4} \cdot 40 = 30.
\]
3. The third term is:
\[
a_3 = \frac{3}{4} \cdot 30 = 22.5 = 22\frac{1}{2}.
\]
4. The fourth term is:
\[
a_4 = \frac{3}{4} \cdot 22.5 = 16.875 = 16\frac{7}{8}.
\]
Thus, the first four terms are:
\[
40,\quad 30,\quad 22\frac{1}{2},\quad 16\frac{7}{8}.
\]
This corresponds to Option B, which is Option 2.
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Simplify this solution Mind Expander
Let’s calculate the first four terms of the sequence starting with \( a_1 = 40 \) and \( a_n = \frac{3}{4} a_{n-1} \). 1. \( a_1 = 40 \) 2. \( a_2 = \frac{3}{4} \times 40 = 30 \) 3. \( a_3 = \frac{3}{4} \times 30 = 22.5 \) (which is \( 22 \frac{1}{2} \)) 4. \( a_4 = \frac{3}{4} \times 22.5 = 16.875 \) (which is \( 16 \frac{7}{8} \)) Therefore, the first four terms are \( 40, 30, 22 \frac{1}{2}, 16 \frac{7}{8} \). The correct answer is Option B: \( 40,30,22 \frac{1}{2}, 16 \frac{7}{8} \).
