Question

Let $a_1, a_2, a_3, \dots$ be a sequence of real numbers. Let $s_n = a_1 + a_2 + \dots + a_n$.
If $2s_n = n(c + a_n)$ for some real number $c$ and for all $n = 1, 2, 3, \dots$, then which one of the following statements is Correct?

• A. $a_1, a_2, a_3, \dots$ is an Arithmetic Progression.
• B. $a_1, 2a_2, 3a_3, \dots$ is an Arithmetic Progression.
• C. $a_1, a_2, a_3, \dots$ is a Geometric Progression.
• D. $a_1, 2a_2, 3a_3, \dots$ is a Geometric Progression.

Hint:

The sum formula of an Arithmetic Progression is \(s_n = \frac{n}{2}(a_1 + a_n)\).
Recognizing this structure immediately when \(c = a_1\) saves significant algebraic effort.

Answer 1

The Correct Option is A

•

Step 1 : Understanding the Question:
We are given a sequence of real numbers \(a_n\) and its sum of the first \(n\) terms, \(s_n\).
We are given the relation \(2s_n = n(c + a_n)\) for all \(n \ge 1\)., where \(c\) is a real constant.
We need to determine the nature of the sequence \(a_n\) or a transformed sequence.

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Step 2 : Key Formula or Approach:
The relationship between the term \(a_n\) and the sum of terms \(s_n\) is given by:
\[ s_n - s_{n-1} = a_n \quad \text{for } n \ge 2 \] For \(n = 1\)., we have \(s_1 = a_1\).
By setting \(n = 1\)., we can find the value of the constant \(c\).
We then substitute this into the recurrence relations to establish the standard properties of the sequence.

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Step 3 : Detailed Explanation:
Let us start by substituting \(n = 1\) into the given relation:
\[ 2s_1 = 1(c + a_1) \] Since \(s_1 = a_1\)., we have:
\[ 2a_1 = c + a_1 \implies c = a_1 \] Substituting \(c = a_1\) back into the general relation:
\[ 2s_n = n(a_1 + a_n) \implies s_n = \frac{n}{2}(a_1 + a_n) \] This is precisely the formula for the sum of the first \(n\) terms of an Arithmetic Progression with the first term \(a_1\) and the \(n\)-th term \(a_n\).
To prove this mathematically, let us use the relation \(s_n - s_{n-1} = a_n\) for \(n \ge 2\):
\[ \frac{n}{2}(a_1 + a_n) - \frac{n-1}{2}(a_1 + a_{n-1}) = a_n \] Multiplying by 2 on both sides:
\[ n(a_1 + a_n) - (n-1)(a_1 + a_{n-1}) = 2a_n \] \[ n a_1 + n a_n - (n-1)a_1 - (n-1)a_{n-1} = 2a_n \] Simplifying the terms:
\[ a_1 + n a_n - (n-1)a_{n-1} = 2a_n \] \[ (n - 2)a_n = (n - 1)a_{n-1} - a_1 \] Let us evaluate this for small values of \(n\):
For \(n = 3\):
\[ (3 - 2)a_3 = (3 - 1)a_2 - a_1 \implies a_3 = 2a_2 - a_1 \implies a_3 - a_2 = a_2 - a_1 \] This shows that the difference between successive terms is constant. Let this common difference be \(d\).
Thus, \(a_2 = a_1 + d\) and \(a_3 = a_1 + 2d\).
For \(n = 4\):
\[ (4 - 2)a_4 = (4 - 1)a_3 - a_1 \implies 2a_4 = 3(a_1 + 2d) - a_1 = 2a_1 + 6d \implies a_4 = a_1 + 3d \] By induction, \(a_n = a_1 + (n-1)d\) for all \(n \ge 1\).
Therefore, the sequence \(a_1, a_2, a_3, \dots\) is an Arithmetic Progression.

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Step 4 : Final Answer:
The sequence \(a_1, a_2, a_3, \dots\) is an Arithmetic Progression.
This corresponds to option (A).