Question

If \( x_1+x_2+x_3+\dots+x_n=pn(n-1) \forall n \in \mathbb{N} \) and \( x_{j+1}-x_j = \text{constant} \, (j=1,2,\dots,n-1) \), then \( (\frac{x_n}{n-1})^2 = \)

• A. \( 4p^2 \)
• B. \( 2p^2 \)
• C. \( \frac{p}{2p-1} \)
• D. \( \frac{p^2+1}{p^2-1} \)

Hint:

For any sequence where the sum is given as a quadratic in \(n\), the common difference \(d\) is twice the coefficient of \(n^2\).

Answer 1

The Correct Option is A

Concept: Since the difference between consecutive terms is constant, the sequence is an Arithmetic Progression (A.P.) with common difference \( d \). The sum of \( n \) terms is \( S_n = \frac{n}{2}(2x_1 + (n-1)d) \).

Step 1: Use the given sum \( S_n = pn(n-1) \).
\( \frac{n}{2}(2x_1 + (n-1)d) = pn(n-1) \Rightarrow 2x_1 + (n-1)d = 2p(n-1) \). This holds for all \( n \), so comparing coefficients: \( d = 2p \) and \( x_1 = 0 \).

Step 2: Find the \(n\)-th term \( x_n \).
\( x_n = x_1 + (n-1)d = 0 + (n-1)(2p) = 2p(n-1) \).

Step 3: Calculate \( (\frac{x_n}{n-1})^2 \).
\( (\frac{2p(n-1)}{n-1})^2 = (2p)^2 = 4p^2 \). 4p^2