Question

What will be the sum of n terms of the following arithmetic progression if the arithmetic progression is: \( (x-y)^2, (x^2+y^2), (x+y)^2, \ldots\ldots\ldots \) up to n terms:

• A. \( n\{(x-y)^2 + (n-1)xy\} \)
• B. \( \{n(x-y)^2 + n(x+y)^2\} \)
• C. \( \frac{n}{2} (x^2+y^2) - (x+y) \)
• D. None of these

Hint:

1. Identify the first term \(a_1\) and common difference \(d\) of the AP. \(a_1 = (x-y)^2\). \(a_2 = x^2+y^2\). Common difference \(d = a_2 - a_1 = (x^2+y^2) - (x^2-2xy+y^2) = 2xy\). (Check: \(a_3 - a_2 = (x+y)^2 - (x^2+y^2) = (x^2+2xy+y^2) - (x^2+y^2) = 2xy\). So it's an AP). 2. Use the sum formula for an AP: \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\). Substitute \(a_1 = (x-y)^2\) and \(d = 2xy\): \(S_n = \frac{n}{2}[2(x-y)^2 + (n-1)(2xy)]\). 3. Factor out 2 from the terms in the square brackets: \(S_n = \frac{n}{2} \cdot 2[(x-y)^2 + (n-1)xy]\). 4. Simplify: \(S_n = n[(x-y)^2 + (n-1)xy]\). This directly matches option (1).

Answer 1

The Correct Option is A

Concept: For an arithmetic progression (AP), the sum of the first \(n\) terms is given by the formula: \( S_n = \frac{n}{2}[2a_1 + (n-1)d] \) where \(a_1\) is the first term and \(d\) is the common difference. First, we need to verify if the given series is indeed an arithmetic progression and find its common difference. Step 1: Expand the terms of the given series Term 1: \(a_1 = (x-y)^2 = x^2 - 2xy + y^2\) Term 2: \(a_2 = x^2 + y^2\) Term 3: \(a_3 = (x+y)^2 = x^2 + 2xy + y^2\) Step 2: Calculate the difference between consecutive terms to find the common difference \(d\) Difference \(d_1 = a_2 - a_1\): \(d_1 = (x^2 + y^2) - (x^2 - 2xy + y^2)\) \(d_1 = x^2 + y^2 - x^2 + 2xy - y^2\) \(d_1 = 2xy\) Difference \(d_2 = a_3 - a_2\): \(d_2 = (x^2 + 2xy + y^2) - (x^2 + y^2)\) \(d_2 = x^2 + 2xy + y^2 - x^2 - y^2\) \(d_2 = 2xy\) Since \(d_1 = d_2 = 2xy\), the series is an arithmetic progression with: First term, \(a_1 = (x-y)^2\) Common difference, \(d = 2xy\) Step 3: Calculate the sum of \(n\) terms (\(S_n\)) Using the formula \( S_n = \frac{n}{2}[2a_1 + (n-1)d] \): Substitute \(a_1 = (x-y)^2\) and \(d = 2xy\): \[ S_n = \frac{n}{2}[2(x-y)^2 + (n-1)(2xy)] \] We can factor out a 2 from the terms inside the main bracket: \[ S_n = \frac{n}{2} \cdot 2 [(x-y)^2 + (n-1)xy] \] \[ S_n = n[(x-y)^2 + (n-1)xy] \] Step 4: Compare the calculated sum \(S_n\) with the given options The calculated sum is \(S_n = n[(x-y)^2 + (n-1)xy]\).
(1) \( n\{(x-y)^2 + (n-1)xy\ \):} This option is identical to our calculated sum \(S_n\).
(2) \( \{n(x-y)^2 + n(x+y)^2\ \):} This does not match. It simplifies to \(n[(x-y)^2 + (x+y)^2] = n[x^2-2xy+y^2 + x^2+2xy+y^2] = n[2x^2+2y^2] = 2n(x^2+y^2)\).
(3) \( \frac{n}{2} (x^2+y^2) - (x+y) \): This does not resemble the sum formula for this AP.
(4) None of these: Since option (1) matches, this is incorrect. Therefore, option (1) is the correct sum.