To solve this problem, we need to verify the condition provided and see which statement about the sequence is correct. We are given a sequence of real numbers \(a_1, a_2, a_3, \dots\) with the condition:
\(2s_n = n(c + a_n)\)
where \(s_n = a_1 + a_2 + \dots + a_n\) and this holds for all positive integers \(n\).
- \(2s_n = n(c + a_n)\) implies \(s_n = \frac{n(c + a_n)}{2}\).
- Substituting for \(s_n\):
- For \(n = 1\): \(s_1 = a_1 = \frac{1}{2}(c + a_1)\) which simplifies to \(a_1 = \frac{1}{2}c + \frac{1}{2}a_1\). Solving gives \(a_1 = c\).
- For \(n = 2\): \(s_2 = a_1 + a_2 = \frac{2}{2}(c + a_2)\) which simplifies to \(s_2 = c + a_2\).
- Generalizing for \(n\): \(s_n = a_1 + a_2 + \dots + a_n = \frac{n}{2}(c + a_n)\).
- Observe \(s_{n-1}\):
- \(s_{n-1} = a_1 + a_2 + \dots + a_{n-1}\),
- then \(s_n = s_{n-1} + a_n\).
- Using the formula:
\(a_1 + a_2 + \dots + a_{n-1} + a_n = \frac{n}{2}(c + a_n)\)
\(a_1 + a_2 + \dots + a_{n-1} = \frac{n-1}{2}(c + a_{n-1})\). - Subtract the two equations:
- \(a_n = \frac{n}{2}(c + a_n) - \frac{n-1}{2}(c + a_{n-1})\)
- Simplifying the subtraction, assuming:
- \(a_n - a_{n-1} = \frac{1}{2}(c + a_n) - a_{n-1}\)\)
- Concluding: This gives the definition of an Arithmetic Progression (AP) where the difference between consecutive terms is constant.
Therefore, based on the derivation, the correct answer is that the sequence \(a_1, a_2, a_3, \dots\) is an Arithmetic Progression.