NDA 2026 — Mathematics PYQ

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Question

NDA 2026 — Mathematics PYQ

NDA | Mathematics | 2026

If $M_{k} = [k k - 1 am p; k - 1 am p; k]$ where $k$ is a natural number, then what is $∣ M_{1} ∣ + ∣ M_{2} ∣ + ∣ M_{3} ∣ + \dots + ∣ M_{50} ∣$ equal to?

यदि $M_{k} = [k k - 1 am p; k - 1 am p; k]$ है, जहाँ k एक धन पूर्णांक है, तो $∣ M_{1} ∣ + ∣ M_{2} ∣ + ∣ M_{3} ∣ + \dots + ∣ M_{50} ∣$ किसके बराबर है ?

Choose the correct answer:

Explanation

To find the sum, we first calculate the determinant of the general matrix $M_{k}$ .

Step 1: Calculate the determinant $∣ M_{k} ∣$

For a $2 \times 2$ matrix $M_{k} = [a c am p; b am p; d]$ , the determinant is given by $a d - b c$ .

$∣ M_{k} ∣ = (k) (k) - (k - 1) (k - 1)$

$∣ M_{k} ∣ = k^{2} - (k^{2} - 2 k + 1)$

$∣ M_{k} ∣ = k^{2} - k^{2} + 2 k - 1$

$∣ M_{k} ∣ = 2 k - 1$

Step 2: Set up the summation

We need to find the sum:

$S = k = 1 \sum 50 ∣ M_{k} ∣ = k = 1 \sum 50 (2 k - 1)$

This is a sum of the first 50 odd numbers. Using the arithmetic series formula $\sum_{k = 1}^{n} (2 k - 1) = n^{2}$ :

$S = 5 0^{2}$

$S = 2500$