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

Step 1: Identify the sequence type.
We are told $x_{j+1} - x_j$ is constant, which is the defining property of an Arithmetic Progression. Call the first term $x_1 = A$ and the common difference $d$.
Step 2: Use a clever shortcut for $x_n$.
A neat trick: the $n$-th term equals the difference of consecutive sums, $x_n = S_n - S_{n-1}$. Here $S_n = pn(n-1)$.
Step 3: Compute $S_n - S_{n-1}$.
We have $S_{n-1} = p(n-1)(n-2)$, so \[ x_n = pn(n-1) - p(n-1)(n-2) = p(n-1)\big[n - (n-2)\big]. \]
Step 4: Simplify.
Inside the bracket, $n - (n-2) = 2$, so \[ x_n = p(n-1)\cdot 2 = 2p(n-1). \]
Step 5: Form the required ratio.
Dividing by $n-1$, \[ \frac{x_n}{n-1} = \frac{2p(n-1)}{n-1} = 2p. \] The $(n-1)$ cancels beautifully.
Step 6: Square it.
Therefore $\left(\dfrac{x_n}{n-1}\right)^2 = (2p)^2 = 4p^2$, which is option (A).
\[ \boxed{4p^2} \]