Question

The sequence \(a_1, a_2, a_3, \dots, a_n, \dots\) is such that \(a_n = \frac{a_{n-1} + a_{n-2}}{2}\) for all \(n \geq 3\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\)?

• A. 12
• B. 16
• C. 20
• D. 24
• E. 28

Hint:

For recursive sequence problems, identify the target term and see which preceding terms you need. Work backwards or forwards from the given information to find the necessary terms step by step.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The problem describes a sequence where each term (from the third term onwards) is the arithmetic mean of the two preceding terms. We are given two terms in the sequence and asked to find a later term.
Step 2: Key Formula or Approach:
The recursive formula for the sequence is given: \[ a_n = \frac{a_{n-1} + a_{n-2}}{2} \] We will use this formula to work our way to the desired term, \(a_6\).
Step 3: Detailed Explanation:
We are given \(a_3 = 4\) and \(a_5 = 20\). Our goal is to find \(a_6\).
The formula for \(a_6\) is: \[ a_6 = \frac{a_5 + a_4}{2} \] We know \(a_5 = 20\), but we don't know \(a_4\). We need to find \(a_4\) first.
We can use the formula for \(a_5\), since we know \(a_5\) and \(a_3\).
\[ a_5 = \frac{a_4 + a_3}{2} \] Now, substitute the known values into this equation: \[ 20 = \frac{a_4 + 4}{2} \] To solve for \(a_4\), we multiply both sides by 2: \[ 40 = a_4 + 4 \] Subtract 4 from both sides: \[ a_4 = 40 - 4 = 36 \] Now that we have the value of \(a_4\), we can find \(a_6\).
Substitute the values of \(a_5\) and \(a_4\) into the formula for \(a_6\): \[ a_6 = \frac{20 + 36}{2} = \frac{56}{2} = 28 \] Step 4: Final Answer:
The value of \(a_6\) is 28.