Find a2, a3, and a4. a1 = 87, a_n = (2/3)a_(n-1) - 1.
Understand the Problem
The question is asking to find the values of a2, a3, and a4 in a sequence, given the first term a1 and a recursive formula for the nth term. The first term is defined as 87, and the recursion relationship involves multiplying the previous term by 2/3 and subtracting 1.
Answer
\( a_2 = 57, a_3 = 37, a_4 = \frac{71}{3} \)
Answer for screen readers
- ( a_2 = 57 )
- ( a_3 = 37 )
- ( a_4 = \frac{71}{3} )
Steps to Solve
- Finding (a_2)
To find the second term (a_2), we use the recursive formula given:
$$a_2 = \frac{2}{3} a_1 - 1$$
Substituting (a_1 = 87):
$$a_2 = \frac{2}{3} \times 87 - 1$$
Calculating this:
$$a_2 = \frac{174}{3} - 1 = 58 - 1 = 57$$
- Finding (a_3)
Next, we find the third term (a_3) using the formula again:
$$a_3 = \frac{2}{3} a_2 - 1$$
Substituting (a_2 = 57):
$$a_3 = \frac{2}{3} \times 57 - 1$$
Calculating this:
$$a_3 = \frac{114}{3} - 1 = 38 - 1 = 37$$
- Finding (a_4)
Finally, we find the fourth term (a_4):
$$a_4 = \frac{2}{3} a_3 - 1$$
Substituting (a_3 = 37):
$$a_4 = \frac{2}{3} \times 37 - 1$$
Calculating this:
$$a_4 = \frac{74}{3} - 1 = \frac{74}{3} - \frac{3}{3} = \frac{71}{3}$$
- ( a_2 = 57 )
- ( a_3 = 37 )
- ( a_4 = \frac{71}{3} )
More Information
The recursive sequence starts with the initial term of 87 and each term is derived from the previous term. Thus, (a_2), (a_3), and (a_4) reflect how quickly values can decrease based on the formula provided.
Tips
- A common mistake is to miscalculate the fractions. Ensure to carefully multiply and divide when working with fractions.
- Also, forgetting to subtract 1 after calculating the fraction can lead to incorrect terms.
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