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

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UpStudy AI · Accepted Answer

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.
The first four terms are $$ 40, 30, 22\frac{1}{2}, 16\frac{7}{8} $$, which is Option B.

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

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